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: 31

Answer

$11(2z-1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ (2z-1)(z+6)-(2z-1)(z-5) .$ 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 $\{ 1,1 \}$ is $ 1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (2z-1),(2z-1) \}$ is $ (2z-1) .$ Hence, the entire expression has $GCF= (2z-1) .$ Factoring the $GCF= (2z-1) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (2z-1) \left( \dfrac{(2z-1)(z+6)}{(2z-1)}-\dfrac{(2z-1)(z-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} (2z-1) \left( (2z-1)^{1-1}(z+6)-(2z-1)^{1-1}(z-5) \right) \\\\= (2z-1) \left( (2z-1)^{0}(z+6)-(2z-1)^{0}(z-5) \right) \\\\= (2z-1) \left( (1)(z+6)-(1)(z-5) \right) \\\\= (2z-1) \left( z+6-z+5 \right) \\\\= (2z-1) \left( 11 \right) \\\\= 11(2z-1) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.