Question

What are the zeros of the function? h(x)=x^3−2x^2−3x

Answer 1

To find the zeroes you make the function equal to zero and then solve. 
$0=x^3-2x^2-3x$
Then factor out the x to get a quadratic inside the parentheses.
 $0=x(x^2-2x-3)$
Factor the inside quadratic formula.
$x(x^2-2x-3)=x((x-3)(x+1))$
Lastly, use the zero product property to solve.
$x*(x-3)*(x+1)=0$
$x=0,x=3,x=-1$