Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 21

Answer

$(p+2)^2(p-2)(p+3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ p^2(p+2)^2+p(p+2)^2-6(p+2)^2 ,$ factor first the $GCF.$ Then factor the resulting trinomial by finding 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 $GCF= (p+2)^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (p+2)^2(p^2+p-6) .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $ 1(-6)=-6 $ and the value of $b$ is $ 1 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-6\}, \{2,-3\}, \\ \{-1,6\}, \{-2,3\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,3 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} (p+2)^2(p^2-2p+3p-6) .\end{array} Grouping the first and second terms and the third and fourth terms of the trinomial, the given expression is equivalent to \begin{array}{l}\require{cancel} (p+2)^2[(p^2-2p)+(3p-6)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} (p+2)^2[p(p-2)+3(p-2)] .\end{array} Factoring the $GCF= (p+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (p+2)^2[(p-2)(p+3)] \\\\= (p+2)^2(p-2)(p+3) .\end{array}
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