Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises - Page 360: 4


$6rt \left( r^2-5rt+3t^2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6r^3t-30r^2t^2+18rt^3 ,$ 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 $\{ 6,-30,18 \}$ is $ 6 $ 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 $\{ r^3t,r^2t^2,rt^3 \}$ is $ rt .$ Hence, the entire expression has $GCF= 6rt .$ Factoring the $GCF= 6rt ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6rt \left( \dfrac{6r^3t}{6rt}-\dfrac{30r^2t^2}{6rt}+\dfrac{18rt^3}{6rt} \right) \\\\= 6rt \left( r^2-5rt+3t^2 \right) .\end{array}
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