1. ## Another fraction prob

I am having some difficulty finding x = 0 for this equation

$\frac{-6x^2-12x}{x^5}=0$

solve for numerator = 0

$-6x^2-12x=0$

then factorize?

$-6x(6x+2)=0$

then for $6x+2=0, x=\frac{-1}{3}$

and x = 0

I don't know if i'm doing it correct

2. Should be -6x(x + 2) = 0

3. $\frac{-6x^2-12x}{x^5}=0$ $\Rightarrow$ $\frac{-6x(x+2)}{x^5}=0$ $\Rightarrow$
$\frac{-6(x+2)}{x^4}=0$

x can't be 0 so i t would be $-6(x+2)=0$ $\Rightarrow$ $x+2=0$ $\Rightarrow$ $x=-2$