Answer
a. See graph
b. The $x$-intercepts are $(-\sqrt2,0)$, $(-1,0)$ and $\left(\sqrt2,0\right)$.
c. The real zeros are $x=-\sqrt2$, $x=-1$ and $x=\sqrt2$.
d. The results from part (c) and those from part (b) are the same.
Work Step by Step
a.
Using a graphing utility, the graph of the function $y=4x^3+4x^2-8x-8$ is as shown.
b.
From the graph, the $x$-intercepts are $(-\sqrt2,0)$, $(-1,0)$ and $\left(\sqrt2,0\right)$.
c.
Set $y=0$:
$$0=4x^3+4x^2-8x-8$$ $$0=4(x^3+x^2-2x-2)$$ $$0=x^3+x^2-2x-2$$ $$0=(x^3+x^2)+(-2x-2)$$ $$0=x^2(x+1)-2(x+1)$$ $$0=(x^2-2)(x+1)$$ $$x^2-2=0$$ $$x^2=2$$ $$x=\pm\sqrt2$$ $$x_1=-\sqrt2$$ $$x_2=-\sqrt2$$ $$x+1=0$$ $$x_3=-1$$
Thus, the real zeros are $x=-\sqrt2$, $x=-1$ and $x=\sqrt2$.
d. The results from part (c) and those from part (b) are the same.
