Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 18

Answer

$(4p+q)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 16p^2-4pq-2q^2 ,$ factor first the $GCF.$ Then, 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. $\bf{\text{Solution Details:}}$ Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 2(8p^2-2pq-q^2) .\end{array} In the trinomial expression above the value of $ac$ is $ 8(-1)=-8 $ and the value of $b$ is $ -2 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-8\}, \{2,-4\}, \{-1,8\}, \{-2,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 2,-4 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(8p^2+2pq-4pq-q^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} 2[(8p^2+2pq)-(4pq+q^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[2p(4p+q)-q(4p+q)] .\end{array} Factoring the $GCF= (4p+q) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(4p+q)(2p-q)] \\\\= 2(4p+q)(2p-q) .\end{array} The missing factor of the given expression is $ (4p+q) .$
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