## Algebra 1

$-\dfrac{(x+3)}{(x+4)}$; $x \ne 3$
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 $=-\dfrac{(x+3)}{(x+4)}$; $x \ne 3$