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: 12

Answer

$\dfrac{6x+1}{3}$

Work Step by Step

Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $ (6x^2-17x-3)\div(3x-9) ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{6x^2-17x-3}{3x-9} \\\\= \dfrac{(6x+1)(x-3)}{3(x-3)} \\\\= \dfrac{(6x+1)(\cancel{x-3})}{3(\cancel{x-3})} \\\\= \dfrac{6x+1}{3} .\end{array}
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