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: 39


$5(m+p)^2 \left( m+p-2-3m^2-6mp-3p^2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 5(m+p)^3-10(m+p)^2-15(m+p)^4 .$ 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 $\{ 5,-10,-15 \}$ is $ 5 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ (m+p)^3,(m+p)^2,(m+p)^4 \}$ is $ (m+p)^2 .$ Hence, the entire expression has $GCF= 5(m+p)^2 .$ Factoring the $GCF= 5(m+p)^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5(m+p)^2 \left( \dfrac{5(m+p)^3}{5(m+p)^2}-\dfrac{10(m+p)^2}{5(m+p)^2}-\dfrac{15(m+p)^4}{5(m+p)^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} 5(m+p)^2 \left( (m+p)^{3-2}-2(m+p)^{2-2}-3(m+p)^{4-2} \right) \\\\= 5(m+p)^2 \left( (m+p)^{1}-2(m+p)^{0}-3(m+p)^{2} \right) \\\\= 5(m+p)^2 \left( (m+p)-2(1)-3(m+p)^{2} \right) .\end{array} Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the expression above is equivalent to \begin{array}{l}\require{cancel} 5(m+p)^2 \left( (m+p)-2(1)-3(m^2+2mp+p^2) \right) .\end{array} Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 5(m+p)^2 \left( m+p-2-3m^2-6mp-3p^2 \right) .\end{array}
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