# Chapter 11 - Rational Expressions and Functions - 11-2 Multiplying and Dividing Rational Expressions - Practice and Problem-Solving Exercises - Page 675: 54

$=\frac{5(2x-5)}{x-5}, with$ $x\ne5$

#### Work Step by Step

Given : $\frac{5x^{2}+10x-15}{5-6x+x^{2}} \div \frac{2x^{2}+7x+3}{4x^{2}-8x-15}$ This becomes : $\frac{5x^{2}+10x-15}{5-6x+x^{2}} \times \frac{4x^{2}-8x-15}{2x^{2}+7x+3}=\frac{5(x-1)(x+3)}{(x-1)(x-5)} \times \frac{(2x+1)(2x-5)}{(x+3)(2x+1)}$ $=\frac{5(2x-5)}{x-5}$ (After dividing out the common factors $(x-1),(x+3)(2x+1)$)

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