## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set: 7

#### Answer

$3(x-4)^2$

#### Work Step by Step

Factoring the $GCF=3$, then the given expression, $3x^2-24x+48$, is equivalent to \begin{array}{l} 3(x^2-8x+16) .\end{array} The two numbers whose product is $ac= 1(16)=16$ and whose sum is $b= -8$ are $\{ -4,-4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $3(x^2-8x+16)$, is \begin{array}{l}\require{cancel} 3(x^2-4x-4x+16) \\\\= 3[(x^2-4x)-(4x-16)] \\\\= 3[x(x-4)-4(x-4)] \\\\= 3[(x-4)(x-4)] \\\\= 3(x-4)^2 .\end{array}

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