Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.4 - Dividing Polynomials: Long Division and Synthetic Division - Exercise Set - Page 370: 9



Work Step by Step

Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $ (2x^2-6x-8)\div(x+1) ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2x^2-6x-8}{x+1} \\\\= \dfrac{2(x^2-3x-4)}{x+1} \\\\= \dfrac{2(x-4)(x+1)}{x+1} \\\\= \dfrac{2(x-4)(\cancel{x+1})}{\cancel{x+1}} \\\\= 2(x-4) \\\\= 2x-8 .\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.