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: 41

Answer

$-r \left( r^{2}-3r-5 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, get the $GCF$ of the given expression, $ -r^3+3r^2+5r .$ 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 $\{ -1,3,5 \}$ is $ 1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ r^3,r^2,r \}$ is $ r .$ Hence, the entire expression has $GCF= r .$ Factoring the $GCF= r ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} r \left( \dfrac{-r^3}{r}+\dfrac{3r^2}{{r}}+\dfrac{5r}{{r}} \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} r \left( -r^{3-1}+3r^{2-1}+5r^{1-1} \right) \\\\= r \left( -r^{2}+3r^{1}+5r^{0} \right) \\\\= r \left( -r^{2}+3r+5(1) \right) \\\\= r \left( -r^{2}+3r+5 \right) .\end{array} Factoring out $-1$ from the second factor results to \begin{array}{l}\require{cancel} -r \left( r^{2}-3r-5 \right) .\end{array}
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