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: 37

Answer

$20z(2z+1)\left( 3z+4 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 15(2z+1)^3+10(2z+1)^2-25(2z+1) .$ 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 $\{ 15,10,-25 \}$ is $ 5 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (2z+1)^3,(2z+1)^2,(2z+1) \}$ is $ (2z+1) .$ Hence, the entire expression has $GCF= 5(2z+1) .$ Factoring the $GCF= 5(2z+1) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5(2z+1) \left( \dfrac{15(2z+1)^3}{5(2z+1)}+\dfrac{10(2z+1)^2}{5(2z+1)}-\dfrac{25(2z+1)}{5(2z+1)} \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} 5(2z+1) \left( 3(2z+1)^{3-1}+2(2z+1)^{2-1}-5(2z+1)^{1-1} \right) \\\\= 5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)^{1}-5(2z+1)^{0} \right) \\\\= 5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)-5(1) \right) \\\\= 5(2z+1) \left( 3(2z+1)^{2}+2(2z+1)-5 \right) .\end{array} Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the expression above is \begin{array}{l}\require{cancel} 5(2z+1) \left( 3(4z^2+4z+1)+2(2z+1)-5 \right) .\end{array} Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 5(2z+1) \left( 12z^2+12z+3+4z+2-5 \right) \\\\= 5(2z+1) \left( 12z^2+(12z+4z)+(3+2-5) \right) \\\\= 5(2z+1) \left( 12z^2+16z \right) .\end{array} Factoring the $GCF=4z$ from the last factor of the expression above results to \begin{array}{l}\require{cancel} 5(2z+1) 4z\left( 3z+4 \right) \\\\= 20z(2z+1)\left( 3z+4 \right) .\end{array}
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