Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 24

Answer

$(3y+z)^2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 9y^2+6yz+z^2 ,$ 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. $\bf{\text{Solution Details:}}$ In the given expression the value of $ac$ is $ 9(1)=9 $ and the value of $b$ is $ 6 .$ The possible pairs of integers whose product is $ac$ 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 $\{ 3,3 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 9y^2+3yz+3yz+z^2 .\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} (9y^2+3yz)+(3yz+z^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3y(3y+z)+z(3y+z) .\end{array} Factoring the $GCF= (3y+z) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3y+z)(3y+z) \\\\= (3y+z)^2 .\end{array}
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