Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises - Page 329: 45


$-2x^2 \left( x^{3}-3x-2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, get the $GCF$ of the given expression, $ -2x^5+6x^3+4x^2 .$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. Finally, factor out the $-1$. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ -2,6,4 \}$ is $ 2 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ x^5,x^3,x^2 \}$ is $ x^2 .$ Hence, the entire expression has $GCF= 2x^2 .$ Factoring the $GCF= 2x^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 2x^2 \left( \dfrac{-2x^5}{2x^2}+\dfrac{6x^3}{2x^2}+\dfrac{4x^2}{2x^2} \right) .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 2x^2 \left( -x^{5-2}+3x^{3-2}+2x^{2-2} \right) \\\\= 2x^2 \left( -x^{3}+3x^{1}+2x^{0} \right) \\\\= 2x^2 \left( -x^{3}+3x+2(1) \right) \\\\= 2x^2 \left( -x^{3}+3x+2 \right) .\end{array} Factoring out $-1$ from the second factor results to \begin{array}{l}\require{cancel} -2x^2 \left( x^{3}-3x-2 \right) .\end{array}
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