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: 40


$3a(p+q) \left( -3a-a^{2}p+a^2q+2p^2+4pq+2q^2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ -9a^2(p+q)-3a^3(p+q)^2+6a(p+q)^3 .$ 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 $\{ -9,-3,6 \}$ is $ 3 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ a^2(p+q),a^3(p+q)^2,a(p+q)^3 \}$ is $ a(p+q) .$ Hence, the entire expression has $GCF= 3a(p+q) .$ Factoring the $GCF= 3a(p+q) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 3a(p+q) \left( \dfrac{-9a^2(p+q)}{3a(p+q)}-\dfrac{3a^3(p+q)^2}{3a(p+q)}+\dfrac{6a(p+q)^3}{3a(p+q)} \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} 3a(p+q) \left( -3a^{2-1}(p+q)^{1-1}-a^{3-1}(p+q)^{2-1}+2a^{1-1}(p+q)^{3-1} \right) \\\\= 3a(p+q) \left( -3a^{1}(p+q)^{0}-a^{2}(p+q)^{1}+2a^{0}(p+q)^{2} \right) \\\\= 3a(p+q) \left( -3a(1)-a^{2}(p+q)+2(1)(p+q)^{2} \right) \\\\= 3a(p+q) \left( -3a-a^{2}(p+q)+2(p+q)^{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} 3a(p+q) \left( -3a-a^{2}(p+q)+2(p^2+2pq+q^2) \right) .\end{array} Using $a(b+c)=ab+ac$ or the Distributive Property, the expression above is equivalent to \begin{array}{l}\require{cancel} 3a(p+q) \left( -3a-a^{2}p+a^2q+2p^2+4pq+2q^2 \right) .\end{array}
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