McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
Continue to next subchapter

Exercise 27 Page 174

Try to rewrite the middle term, bx, as two terms.

(x+2)(x-11)

Practice makes perfect
To factor a trinomial with a leading coefficient of one, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. x^2-9x+(- 22) In this case, we have -22. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign (one positive and one negative.)
Factor Constants Product of Constants
1 and - 22 - 22
-1 and 22 - 22
2 and - 11 - 22
- 2 and 11 - 22

Next, let's consider the coefficient of the linear term. x^2-9x-22 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, -9.

Factors Sum of Factors
1 and - 22 - 21
-1 and 22 21
2 and - 11 -9
-2 and 11 9

We found the factors whose product is - 22 and whose sum is -9. x^2-9x- 22 ⇔ (x+2)(x-11)