## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 57

#### Answer

$\frac{-5}{(x-3)(x+1)(x+2)}$

#### Work Step by Step

Factor $x^{2}+3x+2$: $(x+1)(x+2)$ Factor $x^{2}-2x-3$: $(x+1)(x-3)$ Therefore, the two fractions become: $\frac{1}{(x+1)(x+2)}-\frac{1}{(x+1)(x-3)}$ Find the LCD for the two fractions. This is: $(x+2)(x+1)(x-3)$ Adjust the fractions based on the LCD: $=\frac{x-3}{(x-3)(x+1)(x+2)}-\frac{(x+2)}{(x+2)(x+1)(x-3)}$ Since the denominators are equal, combine the fractions: $=\frac{x-3-(x+2)}{(x-3)(x+1)(x+2)}$ Expand the numerator: $=\frac{x-3-x-2}{(x-3)(x+1)(x+2)}$ Collect like terms: $=\frac{-5}{(x-3)(x+1)(x+2)}$

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