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 - Page 329: 44


$-16y^3 \left( y-4 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, get the $GCF$ of the given expression, $ -16y^4+64y^3 .$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. Finally, factor out the $-1$. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ -16,64 \}$ is $ 16 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ y^4,y^3 \}$ is $ y^3 .$ Hence, the entire expression has $GCF= 16y^3 .$ Factoring the $GCF= 16y^3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 16y^3 \left( \dfrac{-16y^4}{16y^3}+\dfrac{64y^3}{16y^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} 16y^3 \left( -y^{4-3}+4y^{3-3} \right) \\\\= 16y^3 \left( -y^{1}+4y^{0} \right) \\\\= 16y^3 \left( -y+4(1) \right) \\\\= 16y^3 \left( -y+4 \right) .\end{array} Factoring out $-1$ from the second factor results to \begin{array}{l}\require{cancel} -16y^3 \left( y-4 \right) .\end{array}
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