Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set: 80

Answer

$\dfrac{x^{3}-8}{4x-8}=\dfrac{x^{2}+2x+4}{4}$

Work Step by Step

$\dfrac{x^{3}-8}{4x-8}$ Factor the numerator and take out common factor $4$ from the denominator and simplify: $\dfrac{x^{3}-8}{4x-8}=\dfrac{(x-2)(x^{2}+2x+4)}{4(x-2)}=\dfrac{x^{2}+2x+4}{4}$
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