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.

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