Answer
$\dfrac{x^2+7x-1}{(x-3)(x+8)}$
Work Step by Step
Factor $x^2+5x-24$.
With $a=1$ and $c=-24$, $ac=1(-24)=-24$.
The factors of $-24$ whose sum is equal to $b=5$ are $8$ and $-3$.
Rewrite $5x$ as $8x-3x$ to obtain:
$=x^2+8x-3x-24$
Group the terms.
$=(x^2+8x)+(-3x-24)$
Factor each group.
$=x(x+8)-3(x+8)$
Factor out $(x+8)$.
$=(x+8)(x-3)$
Substitute back the factors into the given expression.
$=\dfrac{x}{x-3}-\dfrac{x+1}{(x+8)(x-3)}$
The LCD is $(x+8)(x-3)$.
Make the expressions similar by multiplying $x+8$ to both the numerator and denominator of the first exprssion to obain:
$=\dfrac{x}{x-3}\cdot \dfrac{x+8}{x+8}-\dfrac{x+1}{(x+8)(x-3)}$
Simplify.
$=\dfrac{x(x+8)}{(x-3)(x+8)}-\dfrac{x+1}{(x+8)(x-3)}$
Add numerators because denominators are same.
$=\dfrac{x(x+8)-(x+1)}{(x-3)(x+8)}$
Use distributive property.
$=\dfrac{x^2+8x-x-1}{(x-3)(x+8)}$
Simplify.
$=\dfrac{x^2+7x-1}{(x-3)(x+8)}$
The numerator can no longer be factored.
Hence, the lowest term is:
$\dfrac{x^2+7x-1}{(x-3)(x+8)}$.
