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.
For more, on Factoring Quadratics, go to http://www.math-help-ace.com/Factoring-Trinomials.html
Name : KVLN Age : 47 years old Qualifications : B.Tech., M.S. (from IIT, Madras) Has 14 years of teaching experience. Loves math and chess. Winner of state rank in the mathematical olympiad. University level chess player. Love for math and love for teaching makes him feel more than happy to help. For First-Rate Math Help, go to the author's web site http://www.math-help-ace.com/. It Aims to help to make every one an ace (expert) in math. Explains lucidly math topics for kids and teens with solved examples and exercises. Highlights the salient points and formulas. Helps to develop confidence and desire to continue. Helps to perceive the work as less demanding. Helps to complete their math home work more quickly. Helps to perceive the work as less demanding. Helps to to integrate the current problem with existing knowledge and ideas. Helps to encourage them to reach a solution on their own, with their active mental participation. Helps every student succeed in math by making the journey a pleasant one. The topics you can learn and enjoy include basic algebra and other algebra topics such as Equations, Inequalities, Polynomials, Factoring, Exponents, Logarithms etc. The pleasant journey includes Number Systems and other Numbers topics such as Divisibilty Rules, Prime Factorization, G.C.F., L.C.M., Prime Numbers, Perfect Numbers, Whole Numbers, Integers, Fractions, Decimals, Rational Numbers, Irrational Numbers, Real Numbers etc. Basic operations on numbers such as addition and subtraction including a number of solved examples and Exercises, multiplication including Multiplication Tables and Division including Long Division are lucidly explained. Math Word problems on Addition, Subtraction, Multiplication, Division, G.C.F., L.C.M., Linear Equations in one and Two Variables, Quadratic Equations etc., are lucidly explained.
