Answer
$\displaystyle \frac{x^{2}-6+11}{(x+7)(x+1)^{2}}$
Work Step by Step
Step 1:
Factor each denominator
$x^{2}+8x+7=...$
For $x^{2}+bx+c$, we search for factors of c whose sum is b:
... $+7$ and $+1$ are such factors,
$=(x+7)(x+1)$
$(x+1)^{2}=\qquad$... is already factored.
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 = $(x+7)(x+1)^{2}$
Step 3:
Write each rational expression using the LCM as the denominator. Simplify.
$\displaystyle \frac{2x-3}{(x+7)(x+1)}\cdot\frac{(x+1)}{(x+1)}-\frac{x-2}{(x+1)^{2}}\cdot\frac{(x+7)}{(x+7)}$
$= \displaystyle \frac{2x^{2}+2x-3x-3-(x^{2}+5x-14)}{(x+7)(x+1)^{2}}$
$=\displaystyle \frac{x^{2}-6+11}{(x+7)(x+1)^{2}}$
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 $+11$ add to $-6$, so we leave it as it is.
