## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 21

#### Answer

$(k-3)^2$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $k^2-6k+9 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the given expression, the value of $c$ is $9$ and the value of $b$ is $-6 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,9 \}, \{ 3,3 \}, \{ -1,-9 \}, \{ -3,-3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,-3 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} (k-3)(k-3) \\\\= (k-3)^2 .\end{array}

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