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: 11

Answer

$\dfrac{2x-1}{2}$

Work Step by Step

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