Math Tutorial - Finding the Greatest Common Factor and Factoring Squares

Greatest Common Factor: of two numbers is the result of two numbers being factored into their smaller factors individually and the of all the numbers that are factors, the one that is greatest is the greatest common factor.

Symbol: ( ) means greatest common factor.

For instance, (6,8) means the greatest common factor of 6 and 8.

Example: (6, 8) is,

First, find all the factors of 6. The factors of 6 are all the numbers that when multiplied by another number give us 6. Those are, 1, 2, 3, 6. This is so because:

1*6=6 and 2*3=6

So you see that each number when multiplied by another number gives us 6.

Similarly,

Second, Find the factors of 8.

Those are, 1, 2, 4, 8.

So, now, when we look at the factors of 6 and 8

for 6: 1, 2, 3, 6

for 8: 1, 2, 4, 8

We see that 1, and 2, both appear in the factors of both six and 8:

Now, of the factors that appear in both numbers, that is 1 and 2, the greatest one is 2. *

We use greatest instead of greater because when we speak of bigger numbers, there might be more than just two numbers that are common factors.

Factoring A Difference Of Two Squares

Once students are familiar with the idea of factoring, there are shortcuts to the FOIL method of quadratic equation factoring. One of these is factoring a difference of two squares. A difference of two squares means that we have a monomial multiplied to another monomial to give us a quadratic function where one monomial is a conjugate pair of the other and the middle term in the quadratic form of an equation disappears.

A quadratic equation is represented by ax^2+bx+c, when the "bx" term disappears, we have a difference of two squares.

Example:

(x-3)*(x+3), here (x+3) is a conjugate pair of (x-3), and here is where we get a difference of two squares.

(x-3)(x+3), using the FOIL method, is:

(x)(x)+(x)(3)+(-3)(x)+(-3)(3)=

x^2+3x-3x-9

x^2-9.

Here, we see that since the middle term disappeared, we see that (x-3)(x+3) is simply X^2-(3)^2, the two squares being X and 3, which is why it is called the difference of two squares.

Of course, the same thing would result if we were doing (x+3)(x-3) since multiplication is commutative.

Now, to give more examples:

(x+4)(x-4)= (x^2-4^2)=(x^2-16)

(x+9)(x-9)=(x^2-9^2)=(x^2-81)

And so on, from here on, the patern is too clear and it would be too much repetition to go on.

Numerical Tools For College and University Students

Students attending college or university in a discipline heavy in the physical sciences, for example, science or engineering, frequently make use of several specific numerical routines. Five of the most popular numerical routines are examined below. These types of routines probably cover 90% of the routines a student will use during a typical undergraduate degree. In addition to their popularity among science and engineering programs, these numerical routines are also encountered in many other curriculum. For example, students in first year university who take an Algebra course to satisfy a breadth requirement might need a Simultaneous Equations Solver occasionally - while they are taking the course. Another student might need to apply a Linear Least Squares fit - once, for a specific assignment - when taking an Accounting class. If the students then continue on in their planned majors, say, Political Science or English, they do not use such tools again.

The five routines examined below are presented in response to the following hypothetical question: Which five numerical routines fill most - if not all - of the needs of undergraduate university students? The answer given below presents the most common types of numerical tasks and some of their applications. In addition, several good-quality free tools are named that offer solutions to these types of problems; they provide most of the functionality required by undergraduate students, allowing them to avoid - or at least delay - the expense of purchasing commercial software.

1) Root-finding

Root-finding covers the class of problem in which the zero(s) of an equation cannot be found explicitly.

Consider the Quadratic Equation:

a x^2 + b x + c = 0

a, b, and c are constants, and values of x that satisfy the equation, called the roots or zeros, must be found.

The Quadratic Equation is one example of the class of the problem of finding the roots of polynomial equations which is, in turn, part of the larger class of problem of root-finding. In fact, because the Quadratic Equation is so well-known (students are often introduced to the Quadratic Equation and its solution in Grade 10), root-finding is probably the best-known class of numerical routine.

The van der Waals Equation is another example of a polynomial equation for which roots are often sought:

pV^3 - n(RT + bp)V^2 + n^2 aV - n^3 ab = 0

In this case, values of V that satisfy the equation are sought, and the polynomial is a cubic (the highest power of V is 3). van der Waals Equation is often encountered in chemistry, thermodynamics, and gasdynamics applications.

Kepler's Equation of Elliptical Motion is another equation to which root-finding techniques are applied:

E - e sin(E) = M

In this example, the equation is not a polynomial, but it involves a transcendental function. e and M are known quantities, but there is no way to isolate E on one side of the equation and solve for it explicitly. Consequently, numerical techniques have to be employed. Rearranging the equation as follows turns the problem into one of finding the roots of the equation:

E - e sin(E) - M = 0

These examples are just three equations whose solution requires root-finding; many more equations arise whose solutions can be found only by employing root-finding techniques. Fortunately, the problem of root-finding is a well-developed field of mathematics and computer science. Almost all root-finding algorithms take an iterative approach to computing a solution to a desired degree of accuracy: first, an initial guess is made and checked, then a closer solution is estimated and checked, and this process is repeated until the user-specified level of accuracy is obtained. For example, a user might require four decimal places of accuracy in the solution, so the computer program would stop iterating for a solution once an approximation has been found to four decimal places.

2) Simultaneous Equations

This class of numerical task deals with solving N Equations in N Unknowns. For example, a situation may arise in which it can be mathematically described as a linear (the highest power of x present is 1) system of Three Equations in Three Unknowns:

a11 x1 + a12 x2 + a13 x3 = b1

a21 x1 + a22 x2 + a23 x3 = b2

a31 x1 + a32 x2 + a33 x3 = b3

The aii and bi values are known but the values of xi that satisfy this system of equations must be computed. This task could be accomplished with a pencil, paper, and hand calculator, but it would be tedious. And as systems get larger, the number of computations involved grows fast, introducing the risk of typos or other errors. A system of, say, 10 Equations in 10 Unknowns would keep a person busy for quite a while!

Fortunately, computer programs have been developed that can compute solutions to these systems quickly and accurately. They are usually put in matrix notation:

[A](x) = (b)

where [A] is a square matrix and (x) and (b) are column vectors.

These sorts of systems can arise from almost any field of study. In a course on Linear Algebra such systems will be faced all the time. These systems also arise in electric circuit analysis (i.e. - Mesh Current Analysis), industrial chemistry projects, structural analysis, economics studies, and more. In addition to solving the system for the x values, quantities of the [A] matrix itself are often computed to reveal informative properties (for example, its determinant, eigenvalues, and LU Decomposition).

3) Linear Least-Squares Data Fitting

Linear Least-Squares data fitting is often applied to describe data which includes errors. For example, a curve might be sought for data, but the data may be such that the expected curve does not satisfactorily pass through all the data points. For situations like this, a systematic method is required to produce an approximating function that describes the relationship defined by the data. The approximating function can then be used to interpolate data between the known data points (or to extrapolate outside the range of the known points). Linear least-squares data fitting is one tool available for such situations.

Applications for this class of numerical task arise in almost any field: economics, physics, politics, engineering, chemistry, environmental studies, and many more. For example, say a researcher has collected population data for a country over the past fifty years and would like to define an equation that effectively describes the population growth so that future growth can be extrapolated. Instead of simply looking at the data, and creating a "guesstimate" for an equation--a technique that would vary from one researcher to the next--a systematic and effective way of examining the data is offered by Linear Least- Squares Data Fitting; it offers a systematic approach for determining trends.

4) Interpolation

Interpolation is often used when drawing smooth curves through data, usually data that does not include errors, and provides a systematic technique for computing data values between the known data points (or outside the range of the known data points). For example, a researcher might have (x, y) data points for the following x-values: 1, 2, 3, 4, 5. However, the researcher might need a y-value that corresponds to an x-value of 2.5 or 6.4. The researcher would have to interpolate for the y-value at x = 2.5 (which is within the range of known data values) and extrapolate for the y-value at x = 6.4 (which is outside the range of known data values). Furthermore, the acquisition of the data may require sophisticated equipment that is hard to access, or the data may be very expensive to compute. In these sorts of situations, a systematic method of computing these interpolating data points is required.

Several algorithms exist for this purpose; one such algorithm is a Cubic Spline Interpolation. A Cubic Spline Interpolation creates a smooth curve through known data values by using piecewise third-degree polynomials that pass through all the data values. However, it should be noted that different versions of this algorithm exist, for example, a natural cubic spline interpolation has the second derivatives of the spline polynomial set to zero at the endpoints of the interpolation interval. This means that a graph of the spline outside the known data range is a straight line. Another version of the algorithm forces a "not-a-knot" condition: the second and second-last points are treated as interpolation points rather than knots (i.e. - the interpolating cubics on the first and second sub-intervals are identical, and so are the ones for the last and second last sub-intervals). Applications for spline interpolation include population data gathered over many years, cyclical sales information, and the contour of the shape of an automobile body.

5) Eigenvalues and Eigenvectors

lambda is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied:

[A](v) = lambda (v)

Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue lambda.

Eigenproblems arise in almost all fields of science: structural analysis, computing the modes of vibration of a beam, aeroelasticity and flutter, system stability (structure, aircraft, satellites, etc.), heat transfer, biological systems, population growth, sociology, economics, and statistics. Eigenvalues and eigenvectors are also often used in conjunction with the solution of differential equations. Furthermore, the algorithm behind the Google search engine is also said to treat indexing as an eigenproblem.

Summary

Root-finding, solving Simultaneous Equations, Linear Least-Squares Data-fitting, Interpolation, and the computation of Eigenvalues and Eigenvectors are the most common types of problems faced by students in college and university. Not only are these types of numerical tasks faced by science and engineering students, they also show up throughout a variety of other programs. In addition, two more factors attest to the prevalence of these numerical problems: (i) routines for handling these types of tasks are almost always covered in texts and courses on numerical mathematics, and (ii) algorithms for these mathematical tasks are well-developed and source code for computer programs has been available for decades.

Considering their popularity, readily-available tools that provide solutions to these most common numerical tasks would appeal to a broad range of users. On one hand, some users might need a few routines for one-time or very infrequent use whereas, on the other hand, other users might use a program often, but only one specific routine. In either case, the purchase of a commercial software package is not justified and having free software available is a convenient alternative. In fact, these types of numerical math routines are widely available for free, in a variety of formats, offering a variety of capabilities. Several software packages have been developed for installation on a user's computer, for example, Octave and Scilab, to name two. Others are available as Java applets. And yet more are available as immediate-for-use Javascript web pages; for example, AKiTi.ca offers routines for solving many of these types of problems. The availability of these various numerical routines provides people more options when selecting a tool that best fits their unique needs, especially if these tools include solutions for the most common numerical tasks. The availability of good-quality software tools for working with the most common numerical tasks offers the greatest utility to the greatest number of people.

Introduction to CLEP College Mathematics Examination

In order to determine how skilled people are in mathematics at the undergraduate level, the College Board set up the College Mathematics exam. This is one of the institution's CLEP examinations that are credited by more than three thousand colleges in the United States, which makes it possible to do away with the required entry courses in colleges.

There are a number of topics into which the College Mathematics Exam has been divided in the following manner:

* 25% - Probability & statistics

* 20% - Functions & their graphs

* 20% - Real number systems

* 15% - Other geometry and algebra topics

* 10% - Logic

* 10% - Sets

In the Sets subsection, the topics covered include subsets; union & subsection; venn diagrams; as well as the cartesian product. In the logic subsection, the topics are conditional statements; truth tables; necessary & sufficient conditions; disjunctions, conjunctions, implications, & negations; inverse, converse, & contrapositive; as well as hypotheses, conclusions, & counterexamples.

In the subsection of real number systems, the topics dealt with include odd & even numbers; prime & composite numbers; rational & irrational numbers; factors & divisibility; open & closed intervals; as well as absolute value and order. The topics in the subsection of functions are domain & range; properties & graphs of functions; as well as the composition of functions & inverse functions. Under probability & statistics, the topics include counting problems; simple conditional probability; computing of probabilities of both simple & compound events; median, mean, mode, & range; as well as standard deviation. And in the subsection dealing with additional topics, there are logarithms & exponents; complex numbers; perimeter & area of plane figures; applications from geometry & algebra; parallel and perpendicular lines; pythagorean theorem; properties of circles, rectangles, & triangles; as well as algebraic equations & inequalities.

It requires a maximum of ninety minutes to complete the sixty multiple-choice questions. The score is scaled between 20 and 80, and it takes a minimum of 50 to qualify.

There are more than 1,000 locations worldwide from where the College Mathematics exam is taken, and you may register online at the College Board's website.

How To Help Your Kids Do Algebra

"Mom, I just can't do Algebra."

Of course you'd love to help. But 'Algebra'?

It's a bit outside of a parent's Job Description isn't it?

The good news is you CAN help.

And you don't have to be mathematically minded, either.

All you need is some patience, a little creativity, and being able to see the world through your child's eyes.

Imagine being led into an Egyptian temple, being shown a wall full of weird-looking hieroglyphics, and being asked to translate them.

That's how Algebra feels to some kids.

Your mission is to help your kid understand why we're using hieroglyphics. And then how to use them to solve simple problems.

Here's how you do that:

*** PHASE 1: Get used to the language ***

Algebra uses abbreviations.

It's therefore vital your child is comfortable using abbreviations in daily life.

Explain what abbreviations are and how they're used.

For example, the name Frederick is abbreviated to Fred.

The United States is often abbreviated to the USA or even just the US.

And US states are also abbreviated. New York becomes NY, New Jersey becomes NJ.

Once the basic theory is understood, start introducing abbreviations around the home. There are many ways of doing this. Here are a few ideas to get you started:

~~ Throw a party ~~

Use abbreviations to build up a guest list of invitees.

Bill becomes B.

Marcy becomes M.

And if there's a Jim and a John, use J1 and J2. Or Ji and Jo. Or use surnames to get JW and JS.

~~ Try some home cooking ~~

Teach your child how to make pancakes and give them the recipe as a formula:

2f + 2e + m + s

Where: f=flour, m=milk, s=sugar, e=eggs.

(Okay, I'm no chef, but you get the idea.)

~~ Design a 'healthy eating' schedule ~~

Yes, a little home algebra can even help your family eat more healthily.

Create abbreviations for oranges, apples, bananas, broccoli, celery, etc.

Then plan out your weekly schedule:

Mon: o + a + ba + 2br

Tue: 2ce + p + o + a

Wed: m + 2o

etc.

Once your schedule is drawn up, ask your kid to add up how many of each item you need to buy at the grocery store.

You'd never have guessed a shopping list could make such great algebra training, but it does.

When your child is comfortable with the basic language of Algebra, it's time to move on to the next stage:

*** PHASE 2: Solve some problems ***

Much of Algebra involves finding an unknown value, also known as the "x factor".

The best way to develop problem-solving skills in kids is by playing simple games and puzzles.

Here are two particularly useful games for developing the algebra mind.

~~ "Dollars & Dimes" ~~

The idea is to give your child a formula, and they tell you the amount of money.

So you say something like:

"2d plus n plus i".

This means two dollars plus a nickel plus a dime.

(Notice how the letter 'i' has ingeniously been used to represent a dIme. This is because the letter 'd' has already been reserved for the Dollar.)

The answer should be "Two dollars and fifteen cents."

Play this a few times using several combinations of notes and coins.

You can also tell your child an amount of money, and ask them to give you the formula.

So if you say, "Four dollars and five cents",your child replies: "4d plus 5c".

Ideally they should give you the answer using the least amount of coins possible. So for "fifteen cents", they should say "n + i" or "i + n". Not "15c"!

~~ "Think of a Number" ~~

This is a classic you can play anywhere. Try it on long car journeys.

The objective is to guess the mystery number.

You say: "I've thought of a number, added 3 to it, and the result is 7. What is my number?"

Make the questions as easy as possible to start with. As your child gains confidence, make the questions are little harder.

At some point, say you're going to call the mystery number 'x'. Then ask the question in equation form.

Now don't panic. It's simple.

Suppose your question is: "What number plus 3 makes 17?"

You call the mystery number 'x'. And so your equation is:

x + 3 = 17

You ask your child "if x plus three is seventeen, what's x?"

You can also reverse roles and ask them to give you some puzzles. Most kids enjoy this. And it trains them to think more creatively about algebra too.

*** Summary ***

Just a few subtle changes in the way your child thinks can have profound effects on their results in the math class.

Don't overload your child. Go in very small steps with lots of similar examples to give practice and confidence.

Even the smallest thing may be a stumbling block. Like understanding that 'x' means '1x' (the '1' is not usually written down).

Give lots of praise and reward to create and reinforce the 'feel good' factor.

As in any kind of teaching, it's better to ask lots of questions rather than keep telling someone something.

Take enough little leaps and at some point your kid will experience the "Aha!" moment when the whole algebra thing suddenly clicks into place.

Mastering Algebra - Recognizing Special Products and Their Factors - Part II

Now that we understand some key algebraic terminology, we are prepared to recognize some special products and to be able to factor them accordingly.  Herein we master how to recognize and factor both differences of perfect squares and perfect square trinomials.

Perfect squares are numbers which have square roots which are integers.  Thus 25, 36, and 49 are all perfect squares because their respective square roots are 5, 6, and 7.  When you have an algebraic binomial which is the difference of two squares, we can always factor this expression in a convenient fashion.  Let us examine a specific example.  Take x^2 - y^2.  Here x^2 and y^2 are the squares.  Because we are taking the difference, this expression is aptly named a "difference of two perfect squares."  We can always factor an expression like this as (x - y)(x + y).  If we insert perfect square coefficients in front of the variables, no matter.  We simply factor the expression by taking the square root of the number and the variable, and placing them in that "-", "+" pattern.  For example 49x^2 - 25y^2 is factored as (7x - 5y)(7x + 5y). 

The difference of two perfect squares has a nice application in short-cut arithmetic.  Namely, we can use this method to perform some lightning multiplications.  For example, take any two numbers that differ from a common "ten" by the same amount.  Specifically, take 36 and 44.  Both these numbers differ from 40, the "common ten," by 4 units.  If we want to multiply 36*44, we can get the answer immediately, which is 1584!  How?  Write 36 as 40 - 4 and 44 as 40 + 4.  Then 36*44 = (40 - 4)(40 + 4) which is 40^2 - 4^2, which is 1600 - 16, which is 1584.  Try some others out on your own to see the beauty of this method.

Perfect square trinomials comprise another set of special products.  These expressions are generated by the product of two identical binomials.  To wit, (x + 1)^2 = (x + 1)(x + 1).  When multiplied out, this yields x^2 + 2x + 1.  All perfect square trinomials are formed this way.  These expressions are such that the middle term coefficient is double the product of the square root of the constant term and the coefficient of first term.  In x^2 + 2x + 1, 2 = 2*1*1, 1 being the square root of the constant term 1, and 1 being the square root of the coefficient of x^2, which is also 1.  Whenever we have a trinomial that meets this condition, we can always factor it using the square of the binomial which meets these conditions. 

Take an example to make this clear.  Look at x^2 + 10x + 25.  The square root of 25 is 5.  The square root of 1, the coefficient of x^2 is 1; 2*5*1 is 10, which is the middle term coefficient.  Therefore x^2 + 10x + 25 can be factored as (x + 5)^2.   Take one more.  Look at x^2 - 12x + 36.  The square root of 36 is 6.  The square root of 1, the coefficient of x^2 is 1; 2*6*1 is 12, which is the middle term coefficient with a "-" sign in front.  No problem.  We adjust our product by writing a "-" instead of a "+" and get  x^2 - 12x + 36 = (x - 6)^2.  That's all there is to it.  You can now handle any perfect square trinomial on the planet!

Perfect square trinomials play a very important role in a process known as "completing the square."  This process is so named because geometrically, this involves making a square from a rectangle by "completing the square on the rectangle."  This method allows us to solve quadratic equations quite easily and also gives us the proof of the famous quadratic formula.  So if you're ever wondering where that strange formula on the cover of your algebra book came from, remember that a perfect square trinomial had something to do with it.  Wow, isn't it great when you learn where things come from?  See you next time...

Top 5 Mental Math Methods in the World

Today you can define mental math in various different ways. Some would say, memorizing times table and remembering the solutions can form the part of mental mathematics. Some would say ability to perform simple calculations in your head can be mental mathematics.

The web dictionary defines mental mathematics as "Computing an exact answer without using pencil and paper or other physical aids."

Today there are five methods available to learn and practice mental mathematics.

Let's begin with the first one called 'Learning by Heart' or better known as the rote memorizing method where your teachers ask you to mug up boring multiplication tables. It not only kills the interest of the child in mathematics but also makes sure that he develops hatred towards the subject for the rest of the years he studies it. This system gives its ardent devotee some degree of success initially as he is able to answer easy problems but then when the supposedly bigger application problems come the steam is almost over.

The second one gives you a good degree of success and I would highly recommend it to the younger lot out there. It hails from China and is popular by the name of The Abacus (also known as the Soroban in Japan). An abacus is a calculating tool, often constructed as a wooden frame with beads sliding on wires. With the use of this tool one can perform calculations relating to addition, subtraction, multiplication and division with ease. Gradually one practices with the tool in one's hand and later on when experienced he learns to do it without the tool. This tool is then fitted into the mind mentally and he can then add, subtract multiply and divide in seconds. This tool also enhances a child's concentration levels.

The main drawback of this system is that it focuses only on the 4 mathematical operations. Concepts beyond these operations such as Algebra, Square Roots, Cubes, Squares, Calculus, and Geometry etc cannot be solved using it at all. Also one needs a longer time to be able to fully get a grasp of the system hence you see courses in the abacus stretching to over 2 years which leads the child to boredom and then quitting from the course.

Another Chinese system mainly collected from the book The Nine Chapters on the Mathematical Art lays out an approach to mathematics that centers on finding the most general methods of solving problems. Entries in the book usually take the form of a statement of a problem, followed by the statement of the solution, and an explanation of the procedure that led to the solution.

The methods explained in this system can hardly be termed mental and they lack speed to top it all. The Chinese were definitely the most advanced of the civilization thanks to the Yangtze and Yellow Rivers but if I were to choose out of the two methods given by this culture It would be the abacus.

If wars have a 99.99% downside, sometimes they can have an upside too for they give birth to stories of hope and creativity. The next mental math system was developed during the Second World War in the Nazi Concentration Camp by a Ukrainian Mathematician Jakow Trachtenberg to keep his mind occupied. What resulted is now known as the Trachtenberg Speed System of Mathematics and consists of Rapid Mental Methods of doing Mathematics.

The system consists of a number of readily memorized patterns that allow one to perform arithmetic computations very quickly. It has wider applications than the Abacus and apart from the four basic operation methods it covers Squares and Square Roots.

The method focuses mostly on Multiplication and it even gives patterns for multiplication by particular number say 5,6,7 and even 11 and 12. It then gives a general method for rapid multiplication and a special two finger method. After practicing the method myself I realized that the multiplication was a very applicable mental method but the other methods covered to solve division and square roots were not very friendly and were impossible to be done mentally. I was in search of a much better wholesome method where I could easily perform other operations also. Another drawback of this system was that it too like the abacus failed to have a wider scope i.e to encompass other fields like Algebra, Calculus, Trignometry, Cube Roots etc

A Recommendation by a friend of mine from America introduced me to what is known as the Kumon Math Method. It was founded by a Japanese educator Toru Kumon in 1950s and as of 2007 over 4 million children were studying under the Kumon Method in over 43 different countries.

Students do not work together as a class but progress through the curriculum at their own pace, moving on to the next level when they have achieved mastery of the previous level. This sometimes involves repeating the same set of worksheets until the student achieves a satisfactory score within a specified time limit. In North American Kumon Centers, the mathematics program starts with very basic skills, such as pattern recognition and counting, and progresses to increasingly challenging subjects, such as calculus, probability and statistics. The Kumon Method does not cover geometry as a separate topic but provides sufficient geometry practice to meet the prerequisites for trigonometry, which is covered within the Kumon math program.

I was much impressed with the glamour around Kumon but a glimpse of its curriculum deeply disappointed me. It is not mental at all. It does not offer any special methods to do mathematics and one does not improve one's speed by doing Kumon Math. There is a set curriculum of worksheets which one does till one achieves mastery in the subject. So say for example a sheet on Divison- one would continue to do division by the conventional method till he gets a satisfactory score and then he moves on to a higher level. This certainly doesn't make division any faster and the process is certainly not mental.

A deep thought on the reason of its tremendous popularity in America led me to conclude was the lack of a franchisee business model of the abacus and the Trachtenberg speed system in the 1950s. The franchisee model was essential in taking the course from country to country. This is where Toru Kumon thrived.

Dissapointed with other cultures in the world, my search made me look in my own Indian culture. What I found astonished and amazed me so much that I fell in love with the system and started coaching neighbourhood students in it.

This is easily the World's Fastest Mental Mathematics System called High Speed Vedic Mathematics. It has its roots in Ancient Indian Scriptures called the Vedas meaning 'the fountain head of knowledge'. With it not only you can add, subtract, multiply or divide which is the limiting factor of the abacus but you can also solve complex mathematics such as algebra, geometry, Calculus, and Trigonometry. Some of the most advanced, complex and arduous problems can be solved using the Vedic Maths method with extreme ease.

And all this with just 16 word formulas written in Sanskrit.

High Speed Vedic Mathematics was founded by Swami Sri Bharati Krishna Tirthaji Maharaja who was the Sankaracharya (Monk of the Highest Order) of Govardhan Matha in Puri between 1911 and 1918. They are called "Vedic" as because the sutras are contained in the Atharva Veda - a branch of mathematics and engineering in the Ancient Indian Scriptures.

High Speed Vedic Mathematics is far more systematic, simplified and unified than the conventional system. It is a mental tool for calculation that encourages the development and use of intuition and innovation, while giving the student a lot of flexibility, fun and satisfaction . For your child, it means giving them a competitive edge, a way to optimize their performance and gives them an edge in mathematics and logic that will help them to shine in the classroom and beyond.

Therefore it's direct and easy to implement in schools - a reason behind its enormous popularity among academicians and students. It complements the Mathematics curriculum conventionally taught in schools by acting as a powerful checking tool and goes to save precious time in examinations.

The Trachtenberg Method is often compared to Vedic Mathematics. Infact even some of the multiplication methods are strikingly similar. The Trachtenberg system comes the closest to the Vedic System in comparison and ease of the methods. But the ease and mental solvability of the other method especially division, square roots, cube roots, Algebraic Equations, Trigonometry, Calculus etc clearly gives the Vedic System an edge. Even NASA is said to be using some of this methods applications in the field of artificial intelligence.

There are just 16 Vedic Math sutras or word formulas which one needs to practice in order to be efficient in Vedic Math system. Sutras or Word Math Formulas such as the Vertically and Crosswise, All from Nine and Last from ten helps to solve complex problems with ease and also a single formula can be applied in two or more fields at the same time. The Vertically and Crosswise formula is one such gem by which one can multiply, find squares, solve simultaneous equations and find the determinant of a matrix all at the same time.

If either of these methods is learned at an early age, a student aged 14 can perform lightening fast calculations easily during his examinations and ace through them.

Vedic Mathematics is fast gaining popularity in this millennium. It is being considered as the only mental math system suited for a child as it helps to develop his numerical as well as mental abilities. The methods are new and practical and teach only Mental Rapid Mathematics.

The system does not focus on learning by repetition as in the Kumon Method. The system focuses on improving intelligence by teaching fundamentals and alternate methods. The purpose is not limited to improving performance in the school or tests, but on providing a broader outlook resulting in improved mathematical intelligence and mental sharpness.

To know more about the Vedic Mathematics Sutras - The World's Fastest Mental Math System you can visit http://www.vedicmathsindia.org

This Article is by Gaurav Tekriwal,, The President of the Vedic Maths Forum India who has been conducting High Speed Vedic Math Workshops for the last five years and has trained over seven thousand students across the world in the field. He is the author of the best selling DVD on the subject which contains over 10 hours on the subject. He is an expert in the field and revolutionizes the way children learn math.

Easy Things That Can Raise Your Algebra Test Scores Dramatically!

Get the inside secrets from math teachers... Easy things that can raise your algebra test scores dramatically! By understanding how your teacher grades tests you can significantly raise your math grades even if you have not completely mastered the concepts.

As a middle and high school math teacher I graded thousands of math tests and all along the way I had to determine how many points I would reward a student for their efforts. The process was fairly subjective and I often used my "gut" or best judgment to make a decision on how many points I would award a student for the work on their test. Of course each teacher will grade tests slightly different, but I can assure you most math teachers are looking for certain indicators before they award points.

Students can really help improve their grade if they know what their teacher is looking for and it's much more than the correct answer. How teachers determine grades and how to maximize your scores:

1. Correct answer: Clearly the sure fire way to increase your math test score is answer as many questions with the right answer! So study and be as ready as possible for tests and quizzes. However, let me give you an inside tip to make your study time even more effective. Before your test ask your teacher as many questions about what will be on the test - you will be surprised on how many details you can find out if you pester your teacher for inside information on a test! Knowing more about what will be on a test can focus your efforts to study the right material. Also, you can almost be certain that 70% of the test questions on a test will be a version of a homework problem or lecture problem covered in class-review your notes and homework problems because these problems will be repackaged as the test questions. One last point to stress, even if you answer a question right make sure you show the supporting work. Most math teachers don't trust magically appearing right answers without evidence that you understood how you arrived at the solution. Many teachers will give you NO credit for correct solutions without supporting work! Be smart show off what you know!

2. Simplified solution: Give me the correct answer 1/2 as the fraction 700/1400 and I will take points off! Every math teacher I know will deduct points for correct answers that are not simplified. Examples of correct solutions that are not simplified are unreduced fractions, answers that don't have the proper of units of measure (like an area or volume problem) and variable or number expressions that are not finished. So when it comes to racking up easy points on your tests, finishing and simplifying your solutions go a long way.

3. How much work you show: The best thing a student can do for their grade beyond getting the correct answer is show as much work as possible...even if you don't know what you're doing! Of course the work you show has to be a serious effort at trying to figure out the problem, but if you provide written evidence that you really tried to answer a question you will most likely earn some "charity" points-it all adds up in the end and can make a major difference in your overall grade.

4. How many questions you attempt: Never, never, never leave a question blank-this is the kiss of death for your math test score! I would like to stress a few points here. When taking a test, first invest the majority of time on the problems you feel that you can answer correctly. However, you always need to leave a little time to attempt questions that you doubt you can figure out. I'd like to refer you back to my 3rd point "how much work you show" as what you need to be focusing on with these last ditch problems. Always try your best to leave no question blank. If you show your teacher that you at least tried each problem, often you will discover they will "kick" you a few extra points for your effort.

5. How much understanding you demonstrated: If you show that you understand some parts of a problem your teacher will reward you even if you're not able to complete the problem. For example, let's say you're working on an equation and wrote each step correctly until you were stumped. Guess what...you have earned some points! Now let's take the same problem with the attitude, "I can't do this, so it's a waste of my time" this approach will always yield in no points. The main factor in collecting points on your test score is demonstrating what you know even if you can't get the problem right! Partial credit is what I'm talking about and as teacher I can tell you that partial credit can often be the difference between a B- and an A-. Be smart and get credit for what you know!

6. Whether your work is clear, neat and understandable: Even if your work is correct, if your teacher can't understand it you're at risk for not getting points! Neatness is a struggle for many students, but overcoming sloppiness can result in big points. A way to get on a teacher's good side (which will always improve your score) is to write your work so neatly that it's easy to read and understand. Now, frustrate your teacher by complicated "chicken scratch" and they won't be as motivated to stick around and find points from your answer. Remember that increasing your math test scores will be the sum of a lot of little things you do and writing neatly is definitely one of them!

7. Your attitude and effort: Talk in class, disrupt a lesson, and don't do your homework...I guarantee your teachers will not be extra creative to find points on your tests. Teachers will absolutely help those that help themselves and often that means the difference between a 69% (D) and 70% (C). Of course if you're interested in getting better math grades than you need to act like it in class, your teacher can tell if you're trying. Ask questions, ask for help and do your best-that alone will yield more points and could very well be the difference between passing and failing.

Math Teachers - Cautions With Vocabulary! Your Mental Images May Not Match - Never Assume They Know!

In my early days of teaching--a long time ago--I treated math vocabulary just the same way vocabulary has always be treated in English class. I gave definition and spelling quizzes. The complaints were loud and frequent: "This isn't English class." My favorite was: "What does this have to do with math?" In spite of the complaints, my students knew how to PROPERLY pronounce and spell "commutative," and they had a good understanding of the definition. With all of the time pressures of No Child Left Behind, I, like so many other teachers, was forced to let go of this focus on definitions and spelling. Students went right back to spelling and pronouncing commutative as "communitive" and their understanding of the meaning went right out the window. Commuting to work and back is hardly the same thing as communing with nature, but they never got the concept of the importance of pronunciation, spelling, and definitions.

I recently read a teacher's blog by David Ginsburg in which he was making the point that we math teachers need to work harder on language fluency--especially when we are using terms that have very different meanings in our mathematics class than they have in everyday life. His focus is on elementary school; and my favorite of his examples deals with the word "borrow" in subtraction. Why should we say borrow when it is never going to be returned as the everyday meaning of borrow implies?

The very same issues happen in upper level math. In Algebra, factoring is an extremely important concept. Our students are frequently given instructions like: Factor x^2 - 3x - 28

I know from many years of both teaching and tutoring Algebra that even when students can actually correctly factor this expression, they cannot tell you in words what it actually means "to factor." Similar to Pavlov's dogs salivating to the sound of a bell, many of our students know that when they see the word factor beside a trinomial like x^2 - 3x - 28 they are supposed to write (x - 7)(x + 4). But what does the word "factor" actually mean to them?

They are familiar with this use of the term: (1) washing hands is an important factor in slowing the spread of disease, or (2) teens' failure to use protected sex is a major factor in teen pregnancies. To our students, "factor" implies a reason or a cause. What does that have to do with x^2 - 3x - 28?

Note: In mathematics, to factor means to re-write as multiplication. x^2 - 3x - 28 becomes (x - 7)(x + 4) when written as a multiplication problem.

Another example from Algebra involves another extremely important concept: Solve. We know what it means to solve a puzzle or to solve ones own troubles. How does "put it together," as with a puzzle apply to: Solve the equation x^2 - 3x = 28? Find your own answers sort of applies, but what is an answer to an equation? How do we know when we have an answer to an equation? We simply do not do a good enough job of drilling in the concept that an answer "makes the equation TRUE." We say it once or twice and from then on ASSUME they understand.

In Geometry, we start using protractors to measures angles in degrees. Does that have anything to do with temperature? Are you certain that all of your students are picturing in their minds the same thing for one degree that you are, or do you just assume they are?

On of my biggest frustrations with missing understanding is with the use of the word "value" with respect to fractions. We say that reducing a fraction does not change its "value" or we tell our students to compare 3/4 and 1/2 as if fractions themselves have "value." THEY DO NOT HAVE VALUE--YET. We cannot compare 3/4 and 1/2 until we know 3/4 "of what" and 1/2 "of what." I can almost guarantee you that 3/4 of the money in my pocket is LESS than 1/2 of the money in your pocket. We do not stress enough to our students that to know if 3/4 > 1/2, we must first know that they deal with THE SAME THING. Even on a number line, the reason that 3/4 > 1/2 is because BOTH are fractions of THE SAME UNIT. However, 3/4 of an inch and 1/2 of a foot CANNOT be written 3/4 > 1/2!

I strongly suspect that if you asked Algebra students, "Is 3/4 always greater than 1/2?" the vast majority, if not all, would say "yes." Homework instructions never say "compare these two fractions from a number line" or "assuming these fractions are of the same item, compare things fractions."

We math teachers must be constantly checking for understanding about what students are picturing with terms and definitions. Be honest with them about the possibilities for misunderstanding. If we don't catch these early, the misunderstanding can become practiced and learned and become almost impossible to repair.

Never ASSUME they are picturing what you are. Never ASSUME they know what you think they know. ALWAYS VERIFY!

How to Succeed in Algebra - Factoring Non-Monic Trinomials - Part II

In Part II of this article, we are going to examine a sure-fire way to factor non-monic trinomials. We are going to examine the proof which gives us this powerful method. After you are done with this presentation, you will not only be able to factor any non-monic, you will understand the reason why this method works.

To begin our presentation, let us take an arbitrary non-monic trinomial: to do this we need to introduce the constants a, b, and c. Thus our generic non-monic becomes ax^2 + bx + c, in which a is an integer greater than 1. Our end aim is to reverse FOIL this so that we can factor it into the form (dx + e)*(fx + g), where d, e, f, and g are integers. If you now FOIL this product, you obtain dfx^2 + dgx+ efx + eg, which we can write as dfx^2 + (dg + ef)x + eg, since the common factor of both dg and ef is x.

If we compare this last expression with the original non-monic, we that df must be equal to a; dg + ef must equal b and eg must equal c. If this were not the case, we would have a factored expression which is incorrect. Now this is where some thought is required, so I enjoin you to put your thinking cap on and follow closely. Feel free to reread as necessary to make sure you are following the argument.

Our goal is to determine d,e,f, and g. To do this we use a clever trick. We multiply a and c together. We note that this must equate to dfeg, since a = df and c = eg. Now remember that the middle coefficient, or b, is equal to dg + ef. If we break ac down into all its component factors, and observe which sum is equal to b, then we have found dg and ef. If we then divide a, which is df by dg, we get df/dg = f/g, since the d's cancel; similarly if we divide ef/df = e/d. We now have all four unknown letters solved, namely d,e,f, and g.

Because the last part was cryptic and probably somewhat hard to follow at first blush, I am going to do a specific example which will make this whole argument clear. Let us take the non-monic trinomial 6x^2 + 17x + 5. Let us multiply a*c or 6*5 = 30. Now break down 30 into all of its possible factor pairs: 1 30, 2 15, 3 10, and 5 6. Find the factor pair which when added or subtracted yields the middle coefficient 17. Clearly this is the pair 2 15. Since b = dg + ef, we know that 2 is dg and 15 is ef. If we divide 2 by 6 (which is dg by df) we get 1/3, which tells us that g is 1 and f is 3; similarly if we divide 6 by 15 and reduce to lowest terms, we get 6/15 or 2/5, which tells us that 2 is d and 5 is e. Thus the factorization is (dx + e)*(fx + g) or (2x + 5)*(3x + 1). If you FOIL this, you will see indeed that you get back the original non-monic.

To make this procedure clear, let us do another. Let us factor the non-monic 6x^2 + 41x + 70. We multiply 6*70 = 420. The factor pairs of 420 are 1 420, 2 210, 3 140, 4 105, 5 84, 6 70, 10 42, 14 30, and 20 21. The pair which added or subtracted produces 41 is 20 21. Thus 20 is dg and 21 is ef. When we divide both of these by a which is df and reduce, we have 20/6 = 10/3 = g/f; and 21/6 = 7/2, which is e/d. Thus 6x^2 + 41x + 70 can be factored into the form (dx + e)*(fx + g) = ( 2x + 7)*(3x + 10). Now how's that for sure-fire!

Note that we have not considered here cases of non-monics which involve negative numbers. Such is the case with 6x^2 + x -70. The method, albeit with a slight adjustment, works just as well. Once you have the method down for the cases in which all the numbers are positive, applying it to cases in which either b or c or both are negative, is academic.

Higher Level Mathematical Reasoning

Mathematics presents students with a mode of thinking/reasoning. It includes observation, attention to detail, analysis, synthesis, relevant question asking, and problem solving. It involves some valuable traits like the ability to handle sweat, frustration, dead ends, perseverance, and the discovery, hopefully, that there is wonder, joy, and even some exhilaration at the end. We invite students deeper into or higher up this mode of reasoning year-by-year, subject-by-subject. So what is higher level mathematical reasoning?

A look at some of the approved and adopted texts suggests that a typical answer is, "algebra." Algebra is generally considered to be higher level math thinking for today's school students. Constants, variables, coefficients, expressions, equations, quadratic equations, real, rational, and irrational numbers, and combining like terms... If we can just get middle school, upper and even lower elementary students to start thinking about all of this, we believe that higher level math reasoning is taking place!

But what about geometry students who have already passed Algebra I, but still have not mastered basic number sense concepts involving fractions? For example, I tutored a high school geometry student recently who did not realize that if amount A is half as much as amount B, then amount B must be twice as much as amount A. This student had memorized the formula for determining the measure of an inscribed angle (it is 1/2 the measure of its intercepted arc), and had solved many problems correctly. But when asked to find the measure of the arc when given the measure of the angle, the student was stumped. It seems that for this student, thinking about basic fractional relationships was actually higher level mathematical reasoning-higher than the current level of understanding.

Higher level math reasoning for students is simply whatever the next step is from where they are now. The relationship between 1/2 and twice, or that a group can be both one and many, or that a "1" sitting in the tens column has a different value than a "1" in the ones column are all good higher level math thinking skills for students who do not yet understand those concepts. People generally consider algebra more abstract than arithmetic, because it appears to be less concrete-and therefore it must be the flagship of "higher level mathematical reasoning." But any concept is "abstract" to the student who does not understand it yet!

The critical element is not the level of difficulty of the work, but whether or not the work is being addressed through reasoning. Students who can factor quadratic equations because they have memorized a bunch of rules cannot be said to be applying higher level mathematical reasoning, unless they actually understand why they are doing what they are doing. There is a big difference between "higher level activities" and "higher level mathematical reasoning." When higher level activities are taught through mere memorization or repetitive activities devoid of real understanding, they do not involve any reasoning at all. When lower level activities are taught in ways that make students really think, then those students are involved with higher level mathematical reasoning. And math teaching need not hang its head and feel inferior to other academic disciplines while focusing on these lower level activities.

Another unfortunate answer to what is higher level mathematical reasoning can be seen in the rush to complicate problem sets in textbooks. The geometry book that the student I tutor is using in school, published by a major publisher and state adopted, has outstanding higher level math reasoning problems to solve. I'm having as much fun with some of them as I'm sure that authors and state committee members had. But my student and many in her class are not. There are precious few problems in any section of this book that allow students to develop a confident understanding of the basic concepts and procedures before "higher level math reasoning" is introduced in the form of clever and complicated levels of application.

Rather than leaping to higher level activities that require fluent reasoning that has not yet been developed, the interests of students would be better served if this book (and others like it) presented step-by-step contexts of problems of graduated difficulty-each problem based on the reasoning developed in the previous problem, and preparing students for the next step of reasoning represented in the following problem. The proper function of a math book is to develop mathematical reasoning, not merely to create problems that require its use. By rushing to over-complicate the problems, textbooks unwittingly exclude many students from success, actually thwarting the development of their reasoning and forcing them to rely on mere memorization to cope with their work.

Yes we need to keep earlier concepts and procedures alive by integrating them into the problems in subsequent chapters, and yes students need to explore multiple uses and applications, and yes they need to use all of this to solve mathematical problems and not merely perform arithmetic calculations. I am not arguing against any of this. But enrichment is enriching and higher level mathematical reasoning is only reasoning when students have access to it. We should take as much pride in opening up and developing that next level of higher mathematical reasoning, whatever it may be, as we do in the creative, clever, complicated, and fun problems our mathematical minds conceive. We should remember what it's like for those who are looking to us for guidance. What is higher level mathematical reasoning for them?

Common Mistakes Children Make When Learning Algebra

Hey! I remember you. You were the one thinking about not taking Algebra. I'm glad to see that you've changed your mind. Now, we must make sure you stay successful and don't regret your decision. Remember that the failure rate for first year Algebra is approximately 50% and this isn't because 50% of our students are dumb! There are many different reasons for this miserable state of affairs--some as simple as having weak multiplication skills. (I trust that if your multiplication skills are weak, you will practice those before the school year starts.) What I am going to give you here are some of the big philosophical reasons students fail (sorry about the word "children") and then a list of the really goofy mistakes kids make.

The 5 BIG Issues Behind Student Algebra Failures:

1. Not understanding what Algebra is. Algebra is the set of basic skills that will be needed for all future math classes. Students tend to view Algebra as they do History, meaning that they study for the test and then promptly forget everything they just studied. You simply cannot do this with Algebra because the skills you just learned will keep coming back again, and again. Many of the early skills will form the foundation for other skills. It is a HUGE mistake to think you can study for the test and then just forget everything. The skills of Algebra must be ingrained in your brain. You must learn them the first time you see them and then you must keep reviewing them.

2. Not bothering to memorize properties. I'm not sure why students think the many sets of properties are included in the textbook, but the majority of students consider them a nuisance factor and never bother to understand them or memorize them. This is another very serious mistake. The properties are your guidelines--the rules--of what you can and can't do in different situations--for the rest of mathematics! They must be learned--not forgotten.

3. Sign mistakes. These could almost be listed in the "goofy mistakes" to follow, but sign mistakes happen more frequently than any other type of mistake, so I decided to give them their own category. You have already learned to deal with positive and negative numbers, and what will get added to this are the rules of negation. These rules are not hard. They just require practice. You must put in the necessary practice. And NEVER forget to distribute the negative sign: - (2 - 7) = -2 + 7 = 5.

4. The GOOFY mistakes. Each of these is at first glance somewhat logical, but just a simple check with numbers will show you these are wrong. Learn these rules the correct way and don't do the DUMB versions!

(a) (x + y)^2 is NOT x^2 + y^2. Let x = 2 and y = 3: (2 + 3)^2 = 5^2 = 25 but 2^2 + 3^2 = 4 + 9 = 13 not 25.

(b) 1/(x + y) is NOT 1/x + 1/y. Again, let x = 2 and y = 3. 1/(2 + 3) = 1/5 but 1/2 + 1/3 = 3/6 + 2/6 = 5/6 not 1/5. NOTE: Had the original fraction been 1/xy then it would be OK to separate it as (1/x)(1/y), but you cannot do so with addition on the bottom of the fraction.

A similar situation occurs with radicals: sqr(x +y) is NOT sqr(x) + sqr(y). Let x = 9 and y = 16. sqr(9 +16) = sqr(25) = 5 but sqr(9) + sqr(16) = 3 + 4 =7 NOT 5. Again, multiplication under the radical sign can be separated, but not addition.

(c) Reducing fractions. You may NOT cancel any number next to an + or - sign. ONLY multiplication. (x + y)/(2 + x) is NOT y/2. Again, let x = 2 and y = 3. (2 + 3)/(2 + 2) = 5/4 NOT 3/2. However, (xy)/(2x) DOES reduce to y/2.

(d) Confusing the Rules for Exponents. (x^2)(x^3)(^4) = x^9. ADD exponents when multiplying terms with like bases. (x^4)^5 = x^20. MULTIPLY exponents when raising a power to another power.

(e) Not checking solutions when you've solved an equation. Yes, even you might have made a mistake; or there might have been an extraneous root you need to find. Always check your answers!

(f) Forgetting that x^2 = 9 has two solutions: both 3 and -3. Remember that the degree of the equation states how many solutions there will be. x^2 = 9 is degree 2, so there will be 2 solutions: +3 and -3.

(g) Division by zero is undefined. This becomes a problematic issue with algebraic fractions. For 2/(x - 3), you must remember that denominators may never be zero. So you must look at (x - 3) from the standpoint of are there any numbers that would force x - 3 = 0. The value 3 would do that, so 3 must excluded from any possible solution.

(h) Raising a product to a power. Many students forget about the coefficient. (4x)^2 = 16x^2 NOT 4x^2

5. Not knowing how to study math. Math needs to be read and studied differently that you study history. Math should be read and studied OUT LOUD. You need to make your homework a study tool with problems copied, all work shown, and additional notes to yourself that will help later when studying for the final.

If you understand the big issues and avoid the silly mistakes, you will find Algebra to be surprising easy. That doesn't mean it doesn't require effort; but your efforts will lead to success rather than frustration.

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part V

In Part V of this series, we examine how we solve the last class of factorable quadratics of the form ax^2 + bx - c, in which the b-term is positive and the c-term is negative. Such an example would be x^2 + 4x - 5. This subclass of quadratics are as easily solvable as those of the "bc-negative" class discussed in Part IV of this series.

To show how similar this class is, let's examine x^2 + 4x -5. This is the same quadratic as the first example in Part IV of this series, except the 4x term now is positive instead of negative. As in the last article, we note that 5 is prime and its only factors are 1 and 5. Since the c-term is negative, the 1 and 5 must be of opposite signs. Since the b-term is positive, the larger number must bear the positive sign; otherwise the result of the b-term would be negative. Thus x^2 + 4x - 5 = (x + 5)(x - 1), and the solutions are -5 and 1.

Although this method should be perfectly clear by now, let's reinforce it with two more examples. Let's take the quadratic x^2 + 10x - 24. The factor pairs of 24 are 1-24, 2-12, 3-8, and 4-6. Notice that as the c-term becomes a larger composite number as in this case, generally the number of possible factor pairs increases. When the c-term is a prime number, as in the first example, then the only factor pairs are 1 and the number itself. (By the way the first 10 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29; 2 is the one and only even prime; 1 is not considered a prime number.) Now of the four factor pairs, the only one that can combine to give 10 is 2-12. Since 10 must be positive and 2 and 12 must be of opposite sign, can you guess which must be positive and which negative? Easy enough, right? Thus x^2 + 10x - 24 = (x + 12)(x - 2) and the solutions are -12 and 2.

Finally, we will solve x^2 + 31x - 66. The factor pairs of 66 are 1-66, 2-33, 3-22, and 6-11. The only pair that combines to yield 31 is 2-33. Again, using the argument above this quadratic must factor as (x + 33)(x - 2), and the solutions are -33 and 2. The reader can easily verify that both -33 and 2 are in fact the zeros of this particular quadratic.

After following this series of articles, you are starting to see how quick you can become at algebra once you understand the rules of the game. And this goes for all of algebra: as we break down each component of this subject and apply these techniques, algebra-and indeed math-no longer is a mystery that perplexes, but a mystery that both enriches and enlightens.

See more at Cool Math Site and Cool Algebra Ebooks

Free Algebra Help

Many people can find help for high school or college algebra online when other options are not available. Websites offer free information to help with homework problems. Although many websites provide answers to homework problems, it doesn't always help if the student doesn't understand how to solve the problem on their own since algebra, like most math, depends on learning a set of rules. Any resource that would be helpful must explain how these rules work and how they can be applied to solve homework problems. These rules come in many forms such as formulas, equations, properties, methods and procedures. In addition, it is important to understand key definitions and important math terms. For algebra, in particular, a student has to have a good foundation in arithmetic including fractions and decimals. One of the most important new concepts a student will encounter in algebra is the variable. A variable can be thought of as a special container or placeholder for a number. In general, a variable can represent any number unless its limits are specified.

Although a variable can represent any number, sometimes only one number can be used. Finding this number is called solving the variable. In this case, the variable represents a mathematical solution or answer to a problem. One example of this is the linear equation which is usually represented by two variables, one independent and the other dependent. When a value is inserted into one variable, the other can only have one value which is the solution.

The linear equation is an important concept in algebra. The student is expected to solve linear equations and be able to graph them. The graph of a linear equation is a straight line on the Cartesian coordinate plane. The equation determines the direction the line points which is also known as the slope. It also indicates its location based on where it intersects one of the primary axes that define the plane. Since a line can be determined from just two points, if two points are plotted in the Cartesian coordinate plane also known as the x-y plane, the linear equation that corresponds to these two points can be found. Sometimes a system of linear equations needs to be solved. A system in this case is two or more linear equations that have the same solutions for their variables when solved. On a graph, a system of linear equations is represented by a line for each equation. The intersection of these lines is the graphical solution.

When linear equations have been understood; the next topic in algebra is usually polynomials. A polynomial is a type of function that has two or more terms. It usually means, however, three or more terms in the expression or equation. An expression with just one term is called a monomial while two terms would be a binomial. Each term has either a different set of variables or the same set but the variables are raised to different powers. One important type of polynomial function is the quadratic equation.

The quadratic equation has two forms: expanded form where it is a monomial or polynomial with two or three terms or factored form where it is usually represented by two factors consisting of linear binomials. The graph of the quadratic function is called a parabola and has a cup or bowl shape on the coordinate plane. One of the tasks a student of algebra is expected to do with quadratic equations is to convert them from one form to another. One example is where a product of linear factors is written as an expanded monomial or polynomial. The rule used to go from factored form to expanded form is called the Law of Distribution or the distributive property. The reverse process usually requires factoring a polynomial which can be more complicated. There are many different methods of factoring. Most methods factor a quadratic polynomial by factoring the coefficients. The most effective way is by completing the square or using the quadratic formula.

There are many web sites online that can help math students as well as algebra students to understand the rules and procedures used in algebra. Some of these websites offer free help for math homework problems in algebra.

Algebra 2 Help

Algebra is a branch of Math that deals with the study of the rules of operations and relations. It deals with working with variables and algebraic expressions. Algebra 2 deals with a part of Algebra with topics like Solving Equations and Inequalities, Graphs and Functions, Systems of equations, polynomials, factoring, Fractional Expressions, Exponents, Square roots, Complex Numbers and Quadratic equations. Students often find it difficult to work with Algebra problems and assignments as their knowledge about the basics wouldn't be strong enough. The secret to working with Algebra problems with ease lies there. Students can acquire an upper hand with strong basics.

The online tutoring bridges the learning gap between the students and the subject. They make the process of learning easier and more effective when compared to learning done at school. The Algebra tutors available online are expert in the subject who can help students in understanding the concepts more clearly by giving elaborate explanations and solved examples. As the whole process of learning occurs online, the process becomes an interactive one enabling students to overcome their hitch and clarify their doubts openly from the tutors.

The Algebra 2 help online, as the name suggests, provides a free help with Algebra 2 for students online. Students can get help with understanding the steps involved in the process of solving various problems. They can also get help with solving homework problems in Algebra 2. They can submit their doubts and homework questions online and the tutors will be answering the questions for them. Along with it the tutors would be providing the methodology followed for solving the problem. Hence students can learn solving such homework problems. It is an excellent online resource for the students to utilize for free.

The advantage of having an online help for Algebra 2 is that students can approach the expert online tutors and solve their doubts regarding a particular topic. They get individual attention of the tutors so that they can understand the working more clearly and become confident with solving problems on their own. It is an interactive process where the students can question and understand the logic behind the explanations provided. They can also get innumerable solved examples on a specific formula or concept. By observing them they can try solving practice problems on similar lines. Practicing problems would help them master the topics. This enables them to prepare well for the examinations.

Why Algebra Looks So Hard?

What is algebra and what it does?

Algebra is a branch of mathematics. Algebra deals with the variable activities in our daily lives. It comes after learning the arithmetic in math.

What are the variable activities?

Variable means something that keeps on changing. Variable activities are the activities which don't stay the same over time. They keep on changing, for example; some trends can go up or down, left or right and east or west etc.

For example; the weight of a person never remains the same, it keeps on changing by getting lower or higher every day. Sun doesn't stay at the same spot whole day; it looks changing its position all the day along (it is because of motion of Earth around the Sun though). Share markets keep getting higher and lower on every single moment. A worker's pay changers according to number of hours he/she worked.

Finally, it can be said that algebra is study of activities which keep on changing with time. As we have hundreds of changing activities around us, therefore, algebra is everywhere in our daily lives.

What are the basic concepts need to be learned before starting algebra?

There is some basic Pre-Algebra concepts need to learned before starting algebra. These topics are given below;

Basic addition, subtraction, multiplication and division.
Times tables at least up to times by 10.
Know how to write all the factors of a number, finding greatest common factor (gcf) and least common multiple (LCM).
Fractions and operations on fractions.
Decimals
Integers
Order of operations

What are the main topics learned in algebra?

The main topics in algebra are;

Knowledge about the variables
Know the coefficients and constants
Writing algebraic expressions
Simple linear equations in one variable
Polynomials, degree and type
Operations with the polynomials, such as, adding polynomials, subtracting polynomials, multiplying and dividing polynomials
Rational expressions, Factorization
System of linear equations in two and three variables
Quadratic equations, absolute value equations and inequalities
Patterns in general like sequences and series.

Why is algebra hard?

Algebra is not that hard. One can take it as a challenging course. It looks hard as it is based on very very general terms.

Generic terms mean, for example; let's say you are looking for your younger brother, Vicky, in a crowd. You can call him by saying his name, Vicky aloud. Also your brother is a boy, and you can call him by saying, boy aloud. This time the problem is, there is too many boys in the crowd and may be, all of them start looking at you. So, boy is a general name and Vicky is a proper name.

Now if you recognize your brother from his clothes, height or other appearance you can even find him in the crowd by saying boy aloud. It looks hard, though but once your brother recognizes your voice, he will come to you. Same way you need to call algebra aloud (I mean learning it) and it will come to you like your brother.

Algebra has some rules to follow (as your brother has some unique appearance other than his name) and if you follow these rules, algebra is not hard at all.

So, that's all about basic algebra terminology, I could open in front of you according to my little knowledge of the topic.

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part IV

In Part IV of this series, we examine how we solve factorable quadratics of the form ax^2 - bx - c, in which both the b and c terms are negative. Such an example would be x^2 - 4x - 5. This subclass of quadratics, which we will call the "bc-negatives," are easily solvable using the factoring techniques we learned about in Parts II and III of this series of articles.

Once again, let's get right to the point and solve the quadratic in the first paragraph: x^2 -4x -5. Remember, we are looking for the solutions, or those values of x which when plugged in for this variable make the quadratic zero. This is another way of saying that those values are the zeros of the function, or those values are the ones which make the quadratic vanish. As in Part III, we look to see the factors of the c-term, which in this example is -5. When considering the factor pairs, we will disregard the negative sign and only examine the factors of 5. Since 5 is a prime number, its only factors are 1 and itself.

Remember: the c-term of the quadratic is always formed by multiplying the numbers of the factor pair. Since the c-term is negative, the 1 and 5 must be of opposite sign, since this is the only way we can get a negative product. Thus we must have 1 with a -5 or -1 with a 5. Both (1)(-5) and (-1)(5) = -5. Since the b-term is obtained by adding and since the b-term is negative, the larger number of the factor pair must be negative. If this were not true, then the result would be positive. You can think of this as the larger negative force overpowering the smaller positive force to give a net negative force. You can also relate this back to chemistry in that if you have more negative electrons than positive protons you end up with a net negative charge, and vice versa. (You see how you can relate math back to many other things in the real world.) Consequently, this quadratic must factor as (x - 5)(x + 1) and the solutions are 5 and -1. Once again, you can plug either of these two values back into the original quadratic to show that they make the equation equal to zero.

To see how this method works with another example, let's try x^2 - 4x - 12. The factor pairs of 12 are 1-12, 2-6, and 3-4. The only pair that will combine to give 4 is 2-6. Now we have to make an adjustment for the b-term. We know that (2)(-6) and (-2)(6) = -12. Of these two examples, the greater negative force occurs in the pair (2)(-6) to yield a net negative of -4. This is what we need, and hence x^2 - 4x - 12 = (x + 2)(x - 6), and the answers are -2 and 6. Each of these will make the quadratic vanish or become equal to zero upon substitution.

In the next article, we will look at the final case of factorable quadratics, ones of the form

x^2 + bx - c. With all the tools we now have available, we will be able to dispatch with this last class with the blink of an eye. In fact, if you've followed this series this far, you already have the knowledge to do so now. Stay tuned...

See more at Cool Math Site and Cool Algebra Ebooks

Get Proficient in Algebra Easily With Algebra 2 Homework Help!

Algebra is an important part of mathematics. To learn Algebra 2, the student should be comprehensively meticulous with the basics, while getting educated. The students can get a lot from Algebra 2 homework help program, which can provide the answers to their main queries concerning Algebra. It is necessary for the students to have completed Algebra homework prior to referring this helpful source.

There are so many topics covered in Algebra 2 such as, graphics, exponents, functions, inequalities, equations involving complex numbers and so on. The course syllabus also comprises other topics like rational expressions, polynomials, complex numbers, radicals, solutions of quadratic, cone sections and rational expressions. The student is not acquainted with these terms and these are totally new for him and may seem to be complicated for him. The Algebra 2 homework help acts his personal guide for these complicated topics and problems in mathematics to easily understand them.

If a student wants to appear for any competitive examination or goes for university entrance test, this help is a most useful program for him for his preparation. Students need a solid base in this part of mathematics to go for GED (General Education Development) examination. GED is treated equivalent to High School certificate. GED is must for those students who have not been able to accomplish their regular high school course of studies.

Algebra 2 homework help will prepare them for this. The common examination for university entrance is ACT and SAT. Students preparing for these examinations need not worry to find personal teachers for coaching them to learn Algebra 2. They can get all help to solve the different questions at the resourceful Algebra 2 homework help website.

This help aims at making mathematics simpler for students. Students can overcome all their difficulties in this subject by making use of this program. Algebra 2 homework help program offers guidelines, which make the learning mathematics simpler and more efficiently. However, it requires that the students acquire sharp skills to memorize the things and stand out in this part of mathematics. Algebra 2 homework help guide books furnish students with such methods that build up their memory and generate interest in the subject so that they learn Algebra easily and proficiently.

The students who are weak to memorize the primary things in Algebra can also get helpful tips from Algebra 2 homework help. We cannot conceive to present an elaborated version here; however the basic reference given here can be helpful for students in polishing their memory to quickly respond to all they have studied in Algebra 1.

Algebra 2 homework help is a most successful method for training students to understand and solve problems using the perfect way. For example, in solving an equation, first total number of terms should be looked into and then select and do appropriate type of factorization to arrive at a solution.

The students expand their vital ability to learn and think potentially with Algebra 2 homework help. It also supports to go for higher studies in mathematics as it builds a firm foundation to attempt complex mathematics. It is easy and student friendly. Students who outshine in Algebra have wide openings for distinguished career opportunities. Although, they may not be cognizant about it yet, they can get jobs with high salary packages. Many of them even can join educational institutions and spread the knowledge further. Algebra 2 homework help is key to success for many students.

Pre Algebra - A Prefix to the Study of Algebra

Algebra is an integral part of mathematics. It can be considered, a tough subject, if the basics are not clear. And to attend to this requirement of budding students, the education system of United States introduced the course of Pre Algebra. It is generally taught in schools in between seventh and ninth grades. However for practice it can be started from sixth grade too.

The main aim of Pre Algebra courses is to make the students develop a healthy sense of numerical relationships. The main emphasis of the teachers at this level falls on manipulating integers, fractions, percents, ratio and proportions decimals, working with variables; critical thinking, reading and comprehending graphs and general problem are solving tricks.

Algebra works as a major pillar towards understanding the crucial study of Algebra. It gives the foundation towards the main study. Algebra has now become important in schools and college studies. It hampers the ranking and the grade system to score bad in the subject. Similarly Algebra is important for getting good jobs and establishing a successful life. All the big jobs wants employees to be of good numeric and analytical skill and this can be achieved with the help of Algebra. It develops the sense of logic in human being which helps them to face the future complexities of life. However to attain this level the primary factor is the complete and thorough knowledge of Pre-Algebra. To solve the larger equations you need have a good idea on this.. It is important that you learn the tricks of Pre-Algebra from an experienced teacher who can guide you well on the subject.

In every field of life, Pre-Algebra and Algebra are important. If you become a scientist, there is absolute no way to escape the study of Pre-Algebra. Even if you are working in a small firm or a big construction company you need to master the field of algebra.

A good foundation is important to learn to manipulate equations later.

Algebra for Dopes - It Ain't That Hard - The Quadratic Equation - Part II

As mentioned in Part I of this series, mastering algebra requires little more than clear-cut explanation and some concerted practice. Conquering the diverse aspects of this field necessitates focused thought and willingness to see through the trees to view the forest. Rather than view algebra-which encompasses many different topics and has vast extensions throughout-in one fell swoop, it is much better to view this subject piecemeal. This is the approach that will be taken in this series of articles.

Learning quadratic equations is one of the key plateaus in algebra as this opens the door to higher degree equations and even the calculus. One of the difficulties with this subject area is that the quadratic has many ways of being conquered, so to speak, that is to say, solved. Essentially there are three ways to solve a second-degree, or quadratic equation: one is by factoring; two is by completing the square; and three is by using the famous quadratic formula.

Now this is where it gets a little sticky since each of these three methods is a whole field within itself. Of the three methods, both the formula and the completing the square will always work on any quadratic equation. Factoring will only work on a subclass of the entire class of second-degree equations, and it is this class that we will discuss briefly here. Specifically, we will look at the subclass of factorable quadratics which have terms with all positive coefficients. That is, if we consider the general quadratic ax^2 + bx + c, then a, b, and c will all be positive. For the sake of classification, we will call this subclass of quadratics the "abc positives."

In order to solve an "abc positive" by factoring, it is necessary that the multiplicative factors of the c-term be able to combine additively or subtractively to the b-term. Now that this is starting to sound like a lot of mumbo jumbo, let's break this down and show how simple this really is. Let's do this with a specific example. Take x^2 + 7x + 12. The pair-wise factors of 12 are 1 and 12, 2 and 6, and 3 and 4. When we multiply any of these pair-wise factors we get 12. We look for the pair that combines additively or subtractively to give 7. It is obvious that only 3 and 4 do.

From this, we factor as such: x^2 + 7x + 12 = (x + 3)(x + 4). The solution to this quadratic is now the opposite of 3 and 4, or -3 and -4. That is, if we substitute -3 or -4 into the original quadratic equation, the result will be 0, for this is what it means for a number to solve a quadratic equation: it makes the equation 0 upon substitution. I leave it to the reader to verify that -3 and -4 do indeed do this.

After taking a look at one more, you will now have the secret to solving the entire subclass of abc positives. Take x^2 + 9x + 20. The pair-wise factors of 20 are 1-20, 2-10, and 4-5. The only pair that add to 9 are 4-5. Thus x^2 + 9x + 20 = (x + 4)(x + 5) and the solutions are -4 and -5. Once again, you can confirm that when each of these values is substituted back into the original quadratic, you will get 0.

When the vast subject of algebra is broken down to size, you start to see that it really is not as difficult as some might make it. In fact, you might start to see that it really is quite understandable. After looking at this approach, you come to realize that anyone could master this field. It really is true: algebra really ain't that bad. And once algebra is mastered, you can be thrilled to know that mathematics opens up a magical door which leads to a rich world of discovery.

Algebra for Dopes - It Ain't That Hard - The Quadratic Equation - Part III

In Part III of this series, we examine how we solve factorable quadratics of the form ax^2 - bx + c, in which both the a and c terms are positive, but the b, or middle term, is negative. Such an example would be x^2 - 5x + 6. This class of quadratics, which we will call the "b-negatives" are easily solvable using the factoring techniques we learned about in Part II.

To get right to the chase, let's look at the quadratic in the first paragraph: x^2 -5x + 6. Remember, we are looking for the solutions, or those values of x which when plugged in for this variable make the quadratic zero. This is another way of saying that those values are the zeros of the function, or those values are the ones which make the quadratic vanish. As in Part II, we look to see the factors of the c-term, which in this example is 6. They are: 1-6, and 2-3. Now this is a very good example because of the nature of these two factor pairs. You see, both 1-6 and 2-3 can combine in some way to give 5; for 1 and 6 can be combined subtractively, that is 6-1 = 5; and 2 and 3 can be combined additively, that is 2 + 3 = 5.

To decide which pair works, remember that this article treats the b-negative class of quadratics. Since the c-term, or 6 is obtained from multiplying the factor pair and only two positives or two negatives can produce a positive result, and since the b-term or -5 can only be obtained by adding two negative numbers, the only factor pair that works is 2-3. For if it were 1-6, then -1 and -6 add to -7 not -5. Thus x^2 - 5x + 6 = (x - 3)(x - 2), and the solutions are 3 and 2. The reader can easily verify that either of these values substituted for x in the original quadratic makes the equation vanish.

To see how this method works with one more example, let's try x^2 - 10x + 16. The factor pairs of 16 are 1-16, 2-8, and 4-4. The only pair that will combine to give 10 is 2-8. Since we have a b-negative quadratic we must have both of these numbers as negatives. Therefore, the quadratic factors into (x - 8)(x - 2) and the solutions are 8 and 2. Nothing at all difficult about this.

If you have followed this series, you are starting to see how we take algebra down to size. By "chunking" this subject up into little component pieces, we make it a much more manageable beast. Thus far we have dissected the class of factorable quadratics into two subclasses, each of which is very easy to deal with. Once this view is taken, the rest of algebra becomes just as manageable and much more pleasant.

A B C of Solving Quadratic Equation

An expression of the type a x ² + b x + c = 0 , ( a ≠ 0 ) is called a quadratic equation in the variable x .

The equation a x ² + b x + c = 0 is called the general (or, standard form)

We can solve a quadratic equation by (1) factorization or by (2) applying the formula.

The formula of finding the roots of the quadratic equation is as follows

x = (- b ± √ (b ² - 4 a c) ) / 2 a

Now we will discuss how to solve applied problems. Due to wide variety of applied problems, there is no single solving technique that works in all cases. However the following suggestion proved helpful.

Step: 1 Read the problem carefully and determine what quantity (s) must be found.

Step: 2 assign a variable name to the quantity.

Step: 3 try expressing the problem algebraically, and as well determining which expressions are equal and write the necessary equation (s).

Step: 3 solve the resulting equation (s)

Now go through a simple problem based on formation of quadratic equation and solving

Problem: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58 / 21 . Find the fraction.

Solution:

Let the numerator of the fraction be x (x be an element of I )

Then its denominator is (2x + 1).

So the fraction is x / (2x + 1))

And the reciprocal would be (2 x + 1) / x

According to the problem:

(X / (2 xs + 1)) + ((2 x + 1) / x) = ( 58 / 21 )

(X ² + (2 x + 1) ²) / (x + (2x + 1)) = ( 58 / 21)

[L. C. D is = ( x + ( 2x + 1 ) ]

21 ( x ² + 4 x ² + 4 x + 1 ) = 58 x ( 2x + 1 )

105 x² + 84 x + 21 = 116 x ² + 58 x

11 x ² - 26 x - 21 = 0

11 x ² - (33 - 7) x - 21 = 0 [using middle term factorization]

11 x ²- 33 x + 7 x - 21 = 0

11 x ( x - 3 ) + 7 ( x - 3 ) = 0

(x - 3) (11 x + 7) = 0

Either ( x - 3 ) = 0 , or ( 11x + 7 ) = 0 [ using zero factor theorem ]

x = 3 ,

From, (11x + 7) = 0

We get, x = - (7 / 11)

But x is an integer , neglect x = - (7 / 11)

Take x = 3

So the required fraction, (x / (2x + 1)) = (3 / (2 * 3 + 1))

= (3 / 7)

Now try the following:

The age of a man is twice the square of the age of his son .Eight years hence the age of the man will be 4 years more than thrice the age of his son. Find their present age?

If you cannot solve this problem, you probably need more practice. A good online tutor would be helpful if you plan to master this subject in a short period. Any reasonably good math tutor should do.

Algebra DVDs - An Excellent Learning Tool to Teach Your Kids Algebra

Algebra is the line of demarcation between basic and higher math. Many children panic upon seeing letters and numbers in a mathematical equation. They often run to their parents for help. Parents there is a resource specifically designed to help you teach your kids algebra, Algebra DVDs.

Algebra DVDs can be a parent's best friend. They provide simple step by step instructions on the basics of algebra along with simple exercises allowing parent and child to test their mastery of each topic. The taped lessons use cartoons, songs, and real world examples to reinforce the concepts being taught.

One of the greatest gifts a parent can give a child is to show them your willingness to learn even though you are an adult. This helps children to understand that not knowing something is not bad as long as you are willing to admit you don't know and seek out help.

A parent and child working their way through this teaching tool together can help to strengthen their bond while increasing their mastery of the subject matter.

Mastering mathematics requires a solid understanding of Algebra. Scientist, engineers, business people and industrialists use algebra to solve problems every day. Understanding Algebra gives children more career choices.

Parents can use the guided lessons to help their children understand basic terms like variable, equation, and factors and follow them through the more complex concepts.

One of the great strengths of the Algebra DVDs is the ability to replay them endlessly until you understand a concept. Standard classroom instruction does not allow you to do this. So buy Algebra DVDs, teach your child algebra and watch their confidence grow.

Factoring Trinomials (Quadratics) - Lucid Explanation of the Method With Examples

Consider the product of the two linear expressions (y+a) and (y+b).

(y+a)(y+b) = y(y+b) + a(y+b) = y^2 + by + ay + ab = y^2 + y(a+b) + ab

We can write it as

y^2 + y(a+b) + ab = (y+a)(y+b) .......(i)

Similarly, Consider the product of the two linear expressions (ay+b) and (cy+d).

(ay+b)(cy+d) = ay(cy+d) + b(cy+d) = acy^2 + ady + bcy + bd = acy^2 + y(ad+bc) + bd

We can write it as

acy^2 + y(ad+bc) + bd = (ay+b)(cy+d) .......(ii)

Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factors and Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear Factors

Observing the two Formulas, leads us to the method of Factorization of Quadratic Expressions.

In Equation (i),

the product of coefficient of y^2 and the constant term = ab

and the coefficient of y = a+b = sum of the factors of ab

Similarly, In Equation (ii),

the product of coefficient of y^2 and the constant term = (ac)(bd) = (ad)(bc)

and the coefficient of y = (ad+bc) = sum of the factors of acbd

So, if we can resolve the product of y^2 and the constant term into product of two factors in such a way that their sum is equal to the coefficient of y, then we can factorize the quadratic expression.

We discuss the steps involved in the method and apply it to solve a number of problems.

Method of Factoring Trinomials (Quadratics) :



Step 1 :

Multiply the coefficient of y^2 by the constant term.

Step 2 :

Resolve this product into two factors such that their sum is the coefficient of y

Step 3 :

Rewrite the y term as the sum of two terms with these factors as coefficients.

Step 4 :

Then take the common factor in the first two terms and the last two terms.

Step 5 :

Then take the common factor from the two terms thus formed.

What you get in step 5 is the product of the required two factors.

The method will be clear by the following Solved Examples.

The examples are so chosen that all the models are covered.

Example 1 :

Factorize 9y^2 + 26y + 16

Solution :

Let P = 9y^2 + 26y + 16

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 9 x 16 = 144

Step 2:

We have to express 144 as two factors whose sum = coefficient of x = 26;

144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18; (8 + 18 = 26)

Step 3:

P = 9y^2 + 26y + 16 = 9y^2 + 8y + 18y + 16

Step 4:

P = y(9y + 8) + 2(9y + 8)

Step 5:

P = (9y + 8)(y + 2)

Thus, 9y^2 + 26y + 16 = (9y + 8)(y + 2) Ans.

Example 2 :

Factorize y^2 + 7y - 78

Solution :

Let P = y^2 + 7y - 78

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 1 x -78 = -78

Step 2:

We have to express -78 as two factors whose sum = coefficient of y = 7 ;

-78 = -2 x 39 = -2 x 3 x 13 = -6 x 13; (-6 + 13 = 7)

Step 3:

P = y^2 + 7y - 78 = y^2 - 6y + 13y - 78

Step 4:

P = y(y - 6) + 13(y - 6)

Step 5:

P = (y - 6)(y + 13)

Thus, y^2 + 7y - 78 = (y - 6)(y + 13) Ans.

Example 3 :

Factorize 4y^2 - 5y + 1

Solution :

Let P = 4y^2 - 5y + 1

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 4 x 1 = 4

Step 2:

We have to express 4 as two factors whose sum = coefficient of y = -5 ;

4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]

Step 3:

P = 4y^2 - 5y + 1 = 4y^2 - 4y - y + 1

Step 4:

P = 4y(y - 1) - 1(y - 1)

Step 5:

P = (y - 1)(4y - 1)

Thus, 4y^2 - 5y + 1 = (y - 1)(4y - 1) Ans.

Example 4 :

Factorize 3y^2 - 17y - 20

Solution :

Let P = 3y^2 - 17y - 20

Now, follow the five steps listed above.

Step 1:

Coefficient of y^2 x constant term = 3 x -20 = -60

Step 2:

We have to express -60 as two factors whose sum = coefficient of x = -17 ;

-60 = -20 x 3; (-20 + 3 = -17)

Step 3:

P = 3y^2 - 17y - 20 = 3y^2 - 20y + 3y - 20

Step 4:

P = y(3y - 20) + 1(3y - 20)

Step 5:

P = (3y - 20)(y + 1)

Thus, 3y^2 - 17y - 20 = (3y - 20)(y + 1) Ans.

Example 5 :



Factorize 2 - 5y - 18y^2

Solution :

Let P = 2 - 5y - 18y^2 = -18y^2 - 5y + 2

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = -18 x 2 = -36

Step 2:

We have to express -36 as two factors whose sum = coefficient of y = -5 ;

-36 = -2 x 18 = -2 x 2 x 9 = 4 x -9; [4 + (-9) = -5]

Step 3:

P = -18y^2 - 5y + 2 = -18y^2 + 4y - 9y + 2

Step 4:

P = 2y(-9y + 2) + 1(-9y + 2)

Step 5:

P = (-9y + 2)(2y + 1)

Thus, 2 - 5y - 18y^2 = (-9y + 2)(2y + 1) Ans.

Example 6 :

Factorize (y^2 + y)^2 -18(y^2 + y) + 72

Solution :

Let P = (y^2 + y)^2 -18(y^2 + y) + 72

Put (y^2 + y) = t; Then P = t^2 -18t + 72

Now, follow the five steps listed above.

Step 1:

(Coefficient of t^2) x (constant term) = 1 x 72 = 72

Step 2:

We have to express 72 as two factors whose sum = coefficient of t = -18 ;

72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]

Step 3:

P = t^2 -18t + 72 = t^2 - 12t - 6t + 72

Step 4:

P = t(t - 12) - 6(t - 12)

Step 5:

P = (t - 12)(t - 6)

But t = (y^2 + y);

So, P = (t - 12)(t - 6) = (y^2 + y - 12)(y^2 + y - 6)

In each of these two brackets, there is a Quadratic Polynomial which can be factorised using the five steps above.

y^2 + y - 12 = y^2 + 4y - 3y - 12 = y(y + 4) - 3(y + 4) = (y + 4)(y - 3)

y^2 + y - 6 = y^2 + 3y - 2y - 6 = y(y + 3) - 2(y + 3) = (y + 3)(y - 2)

See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps.

You might have mastered the 5 steps of factorisation by this time, to write directly like this.

Thus,

P = (y^2 + y)^2 -18(y^2 + y) + 72

= (y^2 + y - 12)(y^2 + y - 6)

= (y + 4)(y - 3)(y + 3)(y - 2) Ans.