#### Answer

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

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