College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises - Page 75: 67


$6a^6 (4a+5b)(2a-3b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 48a^8-12a^7b-90a^6b^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. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 48,-12,-90 \}$ is $ 6 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ a^8,a^7,a^6 \}$ is $ a^6 .$ Hence, the entire expression has $GCF= 6a^6 .$ Factoring the $GCF= 6a^6 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6a^6 \left( 8a^2-2ab-15b^2 \right) .\end{array} In the trinomial expression above the value of $ac$ is $ 8(-15)=-120 $ and the value of $b$ is $ -2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{ 10,-12 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6a^6 \left( 8a^2+10ab-12ab-15b^2 \right) .\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} 6a^6 [( 8a^2+10ab)-(12ab+15b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 6a^6 [2a( 4a+5b)-3b(4a+5b)] .\end{array} Factoring the $GCF= (4a+5b) $ of the entire expression above results to \begin{array}{l}\require{cancel} 6a^6 [(4a+5b)(2a-3b)] \\\\= 6a^6 (4a+5b)(2a-3b) .\end{array}
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