Answer
$-\dfrac{(x+3)}{(x+4)}$; $x \ne 3$
Work Step by Step
Given: $\dfrac{9-x^2}{x^2+x-12}$
Need to find the common factors of the given expression.
$\dfrac{9-x^2}{x^2+x-12}=\dfrac{(3^2-x^2)}{(x+4)(x-3)}$
$=\dfrac{(3-x)(3+x)}{(x+4)(x-3)}$
$=-\dfrac{(x-3)(3+x)}{(x+4)(x-3)}$
$=-\dfrac{(x+3)}{(x+4)}$
If we put $x=3$ then numerator becomes zero, which cannot be possible.
After simplification, we get
$=-\dfrac{(x+3)}{(x+4)}$; $x \ne 3$
