Answer
(a) The x-intercepts are (-2,0) and (4,0)
(b) The x-intercepts are (0,0) and (6,0)
Work Step by Step
Real zeros (or the x-intercepts) are found by making the function equal to zero and solving for x. This is made easier if the polynomial function is in the factored form $f(x)=a(x-r_1)(x-r_2)...(x-r_n)$ since we can solve all $(x-r)$'s for zero. In this case, we solve $(x+2) \text{ and } (x-4)^3$ for zero:
$x_1+2=0$
$x_1=-2$
$(x_2-4)^3=0$
$\sqrt[3]{(x_2-4)^3}=\sqrt[3]0$
$x_2-4=0$
$x_2=4$
The x-intercepts of h(x) are (-2,0) and (4,0).
To solve (b), we first find G(x+3) and then do the same procedure again:
$h(x-2)=(x-2+2)(x-2-4)^3$
$h(x-2)=x(x-6)^3$
$x_1=0$
$(x_2-6)^3=0$
$\sqrt[3]{(x_2-6)^3}=\sqrt[3]0$
$x_2-6=0$
$x_2=6$
The x-intercepts of h(x-2) are (0,0) and (6,0).