Question 825381
As per the rational zero theorem,
the possible rational zeros of {{{h(x)}}} are integers factors of {{{20}}} , meaning
-20, -5, -4, -2, -1, 1, 2, 4, 5, and 20.
 
We can try them all, starting by the easiest ones.
Trying 1 and -1, we find {{{x=-1}}} is one zero:
{{{h(1)=1-4-9+16+20=-3-9+16+20=-12+16+20=4+20=24}}}
{{{h(-1)=1+4-9-16+20=5-9-16+20=-4-16+20=-20+20=0}}}
 
Since {{{x=-1}}} is a zero of {{{h(x)}}} , {{{x-(-1)=x+1}}} must be a factor of {{{h(x)}}} .
That means that {{{h(x)}}} divides evenly by {{{x+1}}} , with no remainder.
Dividing {{{h(x)}}} by {{{x+1}}} we get
{{{Q(x)=x^3-5x^2-4x+20}}} , so 
{{{h(x)=(x+1)Q(x)}}} .
(The division can be done by "synthetic division", or whatever method you have been taught in class).
 
As per the rational zero theorem,
the possible rational zeros of {{{Q(x)}}} are integers factors of {{{20}}} , meaning
-20, -5, -4, -2, -1, 1, 2, 4, 5, and 20.
We know that {{{Q(1)<>0}}} because otherwise {{{h(1) would have been zero.
We can try the other possibilities.
{{{Q(-1)=1-5+4+20=-4+4+20=20}}}
We can find that {{{Q(2)=0}}} either by substituting 2 for x,
or by dividing {{{Q(x)}}} by {{{x-2}}} .
{{{Q(2)=2^3-5*2^2-4*2+20=8-5*4-8+20=8-20-8+20=0}}} ,
and dividing we find that
{{{Q(x)=(x-2)(x^2-3x-10)}}} .
 
So,
{{{h(x)=(x+1)Q(x)=(x+1)(x-2)(x^2-3x-10)}}} .
 
Quadratic polynomial {{{x^2-3x-10}}} is easy to factor as
{{{x^2-3x-10=(x+2)(x-5)}}} , so
{{{h(x)=(x+1)Q(x)=(x+1)(x-2)(x^2-3x-10)=(x+1)(x-2)(x+2)(x-5)}}} .
{{{highlight(h(x)=(x+1)(x-2)(x+2)(x-5))}}}
and {{{h(x)=0}}} for
{{{highlight(x=-2)}}} , {{{highlight(x=-1)}}} , {{{highlight(x=2)}}} , and {{{highlight(x=5)}}} .