## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises - Page 329: 26

#### Answer

$16zn^3 \left( zn^{3}+4n^{4}-2z^{2} \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of each term of the given expression, $16z^2n^6+64zn^7-32z^3n^3 .$ 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 $GCF= 16zn^3 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16zn^3 \left( \dfrac{16z^2n^6}{16zn^3}+\dfrac{64zn^7}{16zn^3}-\dfrac{32z^3n^3}{16zn^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} 16zn^3 \left( z^{2-1}n^{6-3}+4z^{1-1}n^{7-3}-2z^{3-1}n^{3-3} \right) \\\\= 16zn^3 \left( z^{1}n^{3}+4z^{0}n^{4}-2z^{2}n^{0} \right) \\\\= 16zn^3 \left( zn^{3}+4(1)n^{4}-2z^{2}(1) \right) \\\\= 16zn^3 \left( zn^{3}+4n^{4}-2z^{2} \right) .\end{array}

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