## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Review Exercises - Page 361: 29

#### Answer

$(3k-2)^2$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $9k^2-12k+4 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $9(4)=36$ and the value of $b$ is $12 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,36 \}, \{ 2,18 \}, \{ 3,12 \}, \{ 4,9 \}, \{ 6,6 \}, \\ \{ -1,-36 \}, \{ -2,-18 \}, \{ -3,-12 \}, \{ -4,-9 \}, \{ -6,-6 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -6,-6 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 9k^2-6k-6k+4 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (9k^2-6k)-(6k-4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3k(3k-2)-2(3k-2) .\end{array} Factoring the $GCF= (3k-2)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3k-2)(3k-2) \\\\= (3k-2)^2 .\end{array}

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