Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 3

Answer

$4qb \left( 3q+2b-5q^2b \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 12q^2b+8qb^2-20q^3b^2 ,$ get the $GCF.$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 12,8,-20 \}$ is $ 4 $ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ q^2b, qb^2,q^3b^2 \}$ is $ qb .$ Hence, the entire expression has $GCF= 4qb .$ Factoring the $GCF= 4qb ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4qb \left( \dfrac{12q^2b}{4qb}+\dfrac{8qb^2}{4qb}-\dfrac{20q^3b^2}{4qb} \right) \\\\= 4qb \left( 3q+2b-5q^2b \right) .\end{array}
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