Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 6

Answer

$(4p-q)(p+q)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 4p^2+3pq-q^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. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $ 4(-1)=-4 $ and the value of $b$ is $ 3 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-4 \}, \{ 2,-2 \}, \\ \{ -1,4 \}, \{ -2,2 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -1,4 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4p^2-pq+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} (4p^2-pq)+(4pq-q^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} p(4p-q)+q(4p-q) .\end{array} Factoring the $GCF= (4p-q) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4p-q)(p+q) .\end{array}
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