Answer
$\displaystyle \frac{-x^{2}+3x+13}{(x-2)(x+1)(x+4)}$
Work Step by Step
Step 1:
Factor each denominator
$x^{2}-x-2=...$
For $x^{2}+bx+c$, we search for factors of c whose sum is b:
... -2 and +1 are such factors,
$=(x-2)(x+1)$
$ x^{2}+2x-8=\qquad$... $+4$ and $-2$ are such factors,
$=(x+4)(x-2)$
Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials.
LCM = $(x-2)(x+1)(x+4)$
Step 3:
Write each rational expression using the LCM as the denominator. Simplify.
$\displaystyle \frac{x+4}{(x-2)(x+1)}\cdot\frac{(x+4)}{(x+4)}-\frac{2x+3}{(x+4)(x-2)}\cdot\frac{(x+1)}{(x+1)}$
$=\displaystyle \frac{x^{2}+8x+16-(2x^{2}+2x+3x+3)}{(x-2)(x+1)(x+4)}$
$=\displaystyle \frac{-x^{2}+3x+13}{(x-2)(x+1)(x+4)}$
For $ax^{2}+bx+c$ in the numerator,
we search for factors of ac whose sum is b, and rewrite the $bx$ term:
... no two factors of $-13$ add to $+3$, so we leave it as it is.
