Intermediate Algebra (12th Edition)

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

Chapter 5 - Mixed Review Exercises - Page 362: 2



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 30a+am-am^2 ,$ factor first the $GCF.$ Then to factor the resulting trinomial, 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 $GCF=a,$ the expression above is equivalent to \begin{array}{l}\require{cancel} a(30+m-m^2) .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $ -1(30)=-30 $ and the value of $b$ is $ 1 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-30\}, \{2,-15\}, \{3,-10\}, \{5,-6\}, \\ \{-1,30\}, \{-2,15\}, \{-3,10\}, \{-5,6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -5,6 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} a(30-5m+6m-m^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} a[(30-5m)+(6m-m^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a[5(6-m)+m(6-m)] .\end{array} Factoring the $GCF= (6-m) $ of the entire expression above results to \begin{array}{l}\require{cancel} a[(6-m)(5+m)] \\\\= a(6-m)(5+m) .\end{array}
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