Answer
$\frac{5}{(x-4)(x+1)(x+3)}$
Work Step by Step
Looking at the numerator:
Find the LCD (i.e. $(x-4)(x+1)$) of the two fractions and then combine:
$\frac{(x-3)(x+1)}{(x-4)(x+1)}-\frac{(x+2)(x-4)}{(x-4)(x+1)}$
$=\frac{x^{2}-2x-3}{(x-4)(x+1)}-\frac{x^{2}-2x-8}{(x-4)(x+1)}$
$=\frac{x^{2}-2x-3-x^{2}+2x+8}{(x-4)(x+1)}$
Collect like terms:
$=\frac{5}{(x-4)(x+1)}$
The original fraction in question becomes: $\frac{\frac{5}{(x-4)(x+1)}}{x+3}$
Apply the fraction rule $\frac{\frac{b}{c}}{a}=\frac{b}{c\times a}$. This then becomes:
$=\frac{5}{(x-4)(x+1)(x+3)}$
