Answer
(a) The x-intercepts are (-3,0) and (2,0)
(b) The x-intercepts are (-6,0) and (-1,0)
Work Step by Step
Real zeros (or the x-intercepts) are found by making the function equal 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 $(x+3)^2 \text{ and } (x-2)$ for zero:
$(x_1+3)^2=0$
$\sqrt{(x_1+3)^2}=\sqrt0$
$x_1+3=0$
$x_1=-3$
$x_2-2=0$
$x_2=0$
The x-intercepts of G are (-3,0) and (2,0).
To solve (b), we first find G(x+3) and then do the same procedure again:
$G(x+3)=(x+3+3)^2(x+3-2)$
$G(x+3)=(x+6)^2(x+1)$
$0=(x+6)^2(x+1)$
$(x_1+6)^2=0$
$\sqrt{(x_1+6)^2}=\sqrt0$
$x_1+6)=0$
$x_1=-6$
$x_2+1=0$
$x_2=-1$
The x-intercepts of G(x+3) are (-6,0) and (-1,0).
