Page 533 - Algebra 1

Example
2
Factoring Trinomials with Negative Leading Coefficients
Factor completely.
a. -x2+x+56
SOLUTION
The leading coefficient is -1, so factor out a -1.
-x2 +x+56
-1(x2 - x - 56) Factor out -1.
Find two numbers that have a product of -56 and a sum of -1. 7 · -8 = -56 and 7 + (-8) = -1
-1(x2 -x-56)=-1(x+7)(x-8)
So,-x2 +x+56=-(x+7)(x-8).
b. -3x3 - 6x2 + 72x SOLUTION
The leading coefficient is -3. The GCF of the terms is 3x. So, factor out -3x, which is -1 · 3x.
-3x3 - 6x2 + 72x
-3x(x2 + 2x - 24) Factor out the GCF and -1.
Find two numbers that have a product of -24 and a sum of 2. -4 · 6 = -24 and -4 + 6 = 2
-3x(x2 +2x-24)=-3x(x-4)(x+6) So,-3x3 -6x2 +72x=-3x(x-4)(x+6).
Some polynomials have more than one variable. Factoring out the GCF first may be helpful.
Factoring Trinomials with Two Variables
Math Reasoning
Analyze Explain why the product of
-(x + 7)(x - 8) and -1(x + 7)(x - 8) are the same.
Example
3
Factor completely.
a. bx4 + 9bx3 + 20bx2 SOLUTION
The GCF is bx2.
bx4 + 9bx3 + 20bx2 bx2(x2 + 9x + 20)
Factor out the GCF.
Online Connection www.SaxonMathResources.com
Find two numbers that have a product of 20 and a sum of 9. 4 · 5 = 20 and 4 + 5 = 9
bx2(x2 +9x+20)=bx2(x+4)(x+5)
So, bx4 + 9bx3 + 20bx2 = bx2(x + 4)(x + 5).
518 Saxon Algebra 1