Answer
$P(x)=(x+4)^{3}$
Zeros: $ \ -4\ \ \ $
Work Step by Step
$P(x)=x^{3}+12x^{2}+48x+64$
$P(-x)=-x^{3}+12x^{2}-48x+64$
Decscart's rule of signs:
P(x) has 0 sign changes $\Rightarrow$ 0 positive zeros.
P(-x) has 3 sign change $\Rightarrow$ 3 or 1 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64$
Testing with synthetic division,
$\left.\begin{array}{l}
-4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &12 & 48 &64\\\hline
&-4 &-32 &-64\\\hline
1& 8 & 16&|\ \ 0\end{array}$
$P(x)=(x+4)(x^{2}+8x+16)$
For the trinomial, recognize a perfect square.
$P(x)=(x+4)(x+4)^{2}=(x+4)^{3}$
Zeros: $ \ -4\ \ \ $
