Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.3 Multiplying and Dividing Rational Expressions - 7.3 Exercises: 40

Answer

$= \frac{(2x-3)(x+7)}{(x-7)(2x+3)}$

Work Step by Step

$\frac{2x^{2}-5x-12}{x^{2}-3x-28} \div \frac{6x^{2}-7x-24}{3x^{2}+13x-56}$ $= \frac{2x^{2}-5x-12}{x^{2}-3x-28} \times \frac{3x^{2}+13x-56}{6x^{2}-7x-24}$ $= \frac{2x^{2}+8x-3x-12}{(x-7)(x+4)} \times \frac{3x^{2}+21x-8x-56}{6x^{2}-16x+9x-24}$ $= \frac{2x(x+4)-3(x+4)}{(x-7)(x+4)} \times \frac{3x(x+7)-8(x+7)}{2x(3x-8)+3(3x-8)}$ $= \frac{(2x-3)(x+4)}{(x-7)(x+4)} \times \frac{(3x-8)(x+7)}{(2x+3)(3x-8)}$ $= \frac{(2x-3)}{(x-7)} \times \frac{(3x-8)(x+7)}{(2x+3)(3x-8)}$ $= \frac{(2x-3)}{(x-7)} \times \frac{(x+7)}{(2x+3)}$ $= \frac{(2x-3)(x+7)}{(x-7)(2x+3)}$
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