Answer
$\dfrac{2(2x^2+5x-2)}{(x-2)(x+2)(x+3)}$
Work Step by Step
Factor $x^2+x-6$.
With $a=1$ and $c=-24$, $ac=1(-24)=-24$.
The factors of $-24$ whose sum is equal to $b=5$ are $3$ and $-1$.
Rewrite $x$ as $3x-2x$.
$=x^2+3x-2x-6$
Group the terms.
$=(x^2+3x)+(-2x-6)$
Factor each group.
$=x(x+3)-2(x+3)$
Factor out $(x+3)$.
$=(x+3)(x-2)$
Factor $x^2-4$:
$=x^2-2^2$
Use special formula $a^2-b^2=(a+b)(a-b)$.
$=(x+2)(x-2)$
Substitute back the factored form of each polynomial into the given expression.
$=\dfrac{4x}{(x+2)(x-2)}-\dfrac{2}{(x+3)(x-2)}$
The LCD is $(x-2)(x+2)(x+3)$.
Make the expressions similar by multiplying $x+3$ to both the numerator and denominator of the first expression, and $x+2$ to the second expression to obtain:
$=\dfrac{4x}{(x+2)(x-2)}\cdot \dfrac{x+3}{x+3}-\dfrac{2}{(x+3)(x-2)}\cdot \dfrac{x+2}{x+2}$
$=\dfrac{4x(x+3)}{(x-2)(x+2)(x+3)}-\dfrac{2(x+2)}{(x-2)(x+2)(x+3)}$
Add numerators and retain the denominator.
$=\dfrac{4x(x+3)-2(x+2)}{(x-2)(x+2)(x+3)}$
Use the distributive property.
$=\dfrac{4x^2+12x-2x-4}{(x-2)(x+2)(x+3)}$
$=\dfrac{4x^2+10x-4}{(x-2)(x+2)(x+3)}$
Factor out $2$ from the numerator.
$=\dfrac{2(2x^2+5x-2)}{(x-2)(x+2)(x+3)}$
The numerator can no longer be factored.
Hence, the lowest term is:
$\dfrac{2(2x^2+5x-2)}{(x-2)(x+2)(x+3)}$
