# Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 45

$- \displaystyle \frac{x^{2}-3x-13}{(x+1)(x-2)(x+4)}$

#### Work Step by Step

1. Find the LCD . 1st denominator = $(x+1)(x-2)$ 2nd denominator = $(x-2)(x+4)$ List factors of the 1st denominator. From each next denominator, add only those factors that do not yet appear in the list. LCD = $(x+1)(x-2)$ 2. Rewrite each rational expression with the the LCDas the denominator = $\displaystyle \frac{(x+4)(x+4)}{(x+1)(x-2)(x+4)}-\frac{(2x+3)(x+1)}{(x-2)(x+4)(x+1)}$ 3. Add or subtract numerators, placing the resulting expression over the LCD. = $\displaystyle \frac{(x+4)(x+4)-(2x+3)(x+1)}{(x+1)(x-2)(x+4)}$ 4. If possible, simplify. = $\displaystyle \frac{x^{2}+8x+16-(2x^{2}+2x+3x+3)}{(x+1)(x-2)(x+4)}$ = $\displaystyle \frac{x^{2}+8x+16-2x^{2}-2x-3x-3}{(x+1)(x-2)(x+4)}$ = $\displaystyle \frac{-x^{2}+3x+13}{(x+1)(x-2)(x+4)}$ = $- \displaystyle \frac{x^{2}-3x-13}{(x+1)(x-2)(x+4)}$

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