College Algebra, Chapter 4, 4.5, Section 4.5, Problem 26

Factor the polynomial $P(x) = x^3 - 64$, and find all its zeros. State the multiplicity of each zero.

To find the zeros of $P$, we set $x^3 - 64 = 0$, by factoring

Using difference of cube, we set


$\begin{equation} \begin{aligned} x^3 - 64 =& 0 \\ \\ (x - 4)(x^2 + 4x + 16) =& 0 \end{aligned} \end{equation}$


By using quadratic formula,


$\begin{equation} \begin{aligned} x =& \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)} \\ \\ =& \frac{-4 \pm \sqrt{-48}}{2} \\ \\ =& \frac{-4 \pm 4 \sqrt{-3}}{2} \\ \\ =& -2 \pm 2 \sqrt{-3} \\ \\ =& -2 \pm 2 \sqrt{3} i \end{aligned} \end{equation}$


So the zeros of $P$ are $4, -2 + \sqrt{3} i$ and $-2 - \sqrt{3} i$.

By factorization,


$\begin{equation} \begin{aligned} P(x) =& x^3 - 64 \\ \\ =& (x - 4) \left[ x - (-2 + \sqrt{3} i) \right] \left[ x - (-2 - \sqrt{3} i) \right] \\ \\ =& (x - 4) \left[ x + (2 - \sqrt{3} i) \right] \left[ x + (2 + \sqrt{3} i) \right] \left[ x + (2 + \sqrt{3} i) \right] \end{aligned} \end{equation}$


We can say that each factors have multiplicity of $1$.