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: 27

Answer

$7ab\left( 2a^{2}b+a-3a^{4}b^{2}+6b^{3} \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 14a^3b^2+7a^2b-21a^5b^3+42ab^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 $( 14,7,-21,42 )$ is $ 7 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $( a^3b^2,a^2b,a^5b^3,ab^4 )$ is $ ab .$ Hence, the entire expression has $ GCF=7ab .$ Factoring the $ GCF=7ab ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 7ab\left( \dfrac{14a^3b^2}{7ab}+\dfrac{7a^2b}{7ab}-\dfrac{21a^5b^3}{7ab}+\dfrac{42ab^4}{7ab} \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} 7ab\left( 2a^{3-1}b^{2-1}+a^{2-1}b^{1-1}-3a^{5-1}b^{3-1}+6a^{1-1}b^{4-1} \right) \\\\= 7ab\left( 2a^{2}b^{1}+a^{1}b^{0}-3a^{4}b^{2}+6a^{0}b^{3} \right) \\\\= 7ab\left( 2a^{2}b+a(1)-3a^{4}b^{2}+6(1)b^{3} \right) \\\\= 7ab\left( 2a^{2}b+a-3a^{4}b^{2}+6b^{3} \right) .\end{array}
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