Answer
$\displaystyle \frac{x^{2}+7x-1}{(x-3)(x+8)}$
Work Step by Step
Step 1:
Factor each denominator
For $x^{2}+bx+c$, we search for factors of c whose sum is b:
$ x^{2}+5x-24=\qquad$ ... +8 and -3
$=(x+8)(x-3)$
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 = $(x+8)(x-3)$
Step 3:
Write each rational expression using the LCM as the denominator. Simplify.
$\displaystyle \frac{x(x+8)}{(x-3)(x+8)}-\frac{x+1}{(x+8)(x-3)}=$
$=\displaystyle \frac{x(x+8)-(x+1)}{(x-3)(x+8)}$
$=\displaystyle \frac{x^{2}+8x-x-1}{(x-3)(x+8)}$
$=\displaystyle \frac{x^{2}+7x-1}{(x-3)(x+8)}$
... no two factors of -1 add to seven, so we leave the numerator as it is.
