College Algebra, Chapter 4, 4.5, Section 4.5, Problem 34

Factor the polynomial $P(x) = x^6 + 16 x^3 + 64$ and find all its zeros. State the multiplicity of each zero
To find the zeros of $P$, we set $x^6 + 16 x^3 + 64 = 0$, then,

$\begin{equation} \begin{aligned} x^6 + 16x^3 + 64 &= 0 \\ \\ w^2 + 16w + 64 &= 0 && \text{Let } w =x^3\\ \\ (w + 8)^2 &= 0 && \text{Factor}\\ \\ (x^3 + 8)^2 &= 0 && \text{Substitute } w = x^3 \end{aligned} \end{equation}$

The possible rational zeros are the factors of $8$, which are $\pm1, \pm2, \pm4$ and $\pm 8$. Then by using synthetic division and by trial and error



Thus, $P(x) = (x^3 + 8)^2 = \left[ (x+2)(x^2-2x+4) \right]^2$
Then, by using quadratic formula to get the complex zeros,

$\begin{equation} \begin{aligned} x &= \frac{-(-2)\pm\sqrt{(-2)^2 -4(1)(4)}}{2(1)}\\ \\ &= \frac{2\pm\sqrt{-12}}{2}\\ \\ &= \frac{2}{2 \pm 2\sqrt{-3}}\\ \\ &= 1 \pm \sqrt{3}i \end{aligned} \end{equation}$

Hence, the zeros of $P$ are $-2, 1 + \sqrt{3}i$ and $1 -\sqrt{3}i$. Each zeros has a multiplicity of $2$.