Answer
$6,-6$ and $11$
Work Step by Step
The given function is
$\Rightarrow h(x)=(x^2-36)(x-11)$
Substitute $0$ for $h(x)$.
$\Rightarrow 0=(x^2-36)(x-11)$
$\Rightarrow 0=(x^2-6^2)(x-11)$
Use difference of two squares pattern.
$\Rightarrow 0=(x-6)(x+6)(x-11)$
Use zero product property.
$\Rightarrow x-6=0$ or $x+6=0$ or $x-11=0$
Solve for $x$.
$\Rightarrow x=6$ or $x=-6$ or $x=11$
Hence, the zeros of the function are $6,-6$ and $11$.
