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: 24

Answer

$3yz^3\left( 5y^{2}-9yz+1 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of each term of the given expression, $ 15y^3z^3-27y^2z^4+3yz^3 .$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ Using the $GCF= 3yz^3 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3yz^3\left( \dfrac{15y^3z^3}{3yz^3}-\dfrac{27y^2z^4}{3yz^3}+\dfrac{3yz^3}{3yz^3} \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} 3yz^3\left( 5y^{3-1}z^{3-3}-9y^{2-1}z^{4-3}+y^{1-1}z^{3-3} \right) \\\\= 3yz^3\left( 5y^{2}z^{0}-9y^{1}z^{1}+y^{0}z^{0} \right) \\\\= 3yz^3\left( 5y^{2}(1)-9yz+1(1) \right) \\\\= 3yz^3\left( 5y^{2}-9yz+1 \right) .\end{array}
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