## Algebra 2 (1st Edition)

$f(x)=2(x+2)(x+1)(x-2)$
Here, we have $f(x)=a (x-(-2))(x-(-1))(x-2)$ or, $f(x)=a (x+2)(x+1)(x-2)$ When $x=0$, then we have $-8=a (0+2)(0+1)(0-2)$ or, $-4a=-8$ This gives: $a=2$ Thus, we have $f(x)=2(x+2)(x+1)(x-2)$