College Algebra (11th Edition)

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

Chapter R - Test - Page 78: 17



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 24m^3-14m^2-24m ,$ 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 $\{ 24,-14,-24 \}$ is $ 2 $ 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 $\{ m^3,m^2,m \}$ is $ m .$ Hence, the entire expression has $GCF= 2m .$ Factoring the $GCF= 2m ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2m \left( \dfrac{24m^3}{2m}-\dfrac{14m^2}{2m}-\dfrac{24m}{2m} \right) \\\\= 2m \left( 12m^2-7m-12 \right) .\end{array} In the trinomial expression above the value of $ac$ is $ 12(-12)=-144 $ and the value of $b$ is $ -7 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 9,-16 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2m \left( 12m^2+9m-16m-12 \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} 2m [(12m^2+9m)-(16m+12)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2m [3m(4m+3)-4(4m+3)] .\end{array} Factoring the $GCF= (4m+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2m [(4m+3)(3m-4)] \\\\= 2m(4m+3)(3m-4) .\end{array}
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