## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{5}{(x-4)(x+1)(x+3)}$
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: $=\frac{5}{(x-4)(x+1)(x+3)}$