## Algebra 1: Common Core (15th Edition)

$6(x-2), where$ $x\ne-2$
Given : $\frac{6x^{2}-24}{x+2}= \frac{6(x^{2}-4)}{x+2}=\frac{6(x+2)(x-2)}{x+2}$ (Since $(a^{2}-b^{2})=(a+b)(a-b)$) $= 6(x-2)$ (After dividing out the common factor (x+2)) $6(x-2), where$ $x\ne-2$