Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$ 
Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$

I am having difficulty factoring the whole equation so that I can equate each factor to $0$ one by one and get the roots accordingly.

Divide both sides by $x^4$ and change variable to $y = \frac{x-1}{x^2}$, we have
$$\begin{align}1 = 5y(1-y)
\iff & 4y(y-1) = -\frac45 \iff 
(2y-1)^2 = \frac15\\
\implies & y = \frac12\left(1 \pm \frac{1}{\sqrt{5}}\right) = \frac{\sqrt{5}\pm 1}{2\sqrt{5}} = \frac{2}{5 \mp \sqrt{5}}
\end{align}
$$
Substitute $y$ back by $\frac{x-1}{x^2}$, we get
$$
 x^2 - \frac{5\mp\sqrt{5}}{2} (x - 1) = 0
\iff \left(x - \frac{5\mp\sqrt{5}}{4}\right)^2 = \frac{5\mp\sqrt{5}}{2} - 
\left(\frac{5\mp\sqrt{5}}{4}\right)^2
= -\frac{5 \pm \sqrt{5}}{8}
$$
This leads to
$\displaystyle\;x = \frac{5 - \epsilon \sqrt{5} \pm i \sqrt{2(5 + \epsilon \sqrt{5})}}{4}$ where $\epsilon = \pm 1$.
If one insists on factoring the equation first, one can treat $x^2$ and $(x-1)$ as two seperate units and rearrange them:
$$\begin{align}
 & x^4 = 5(x-1)(x^2 - x + 1)\\
\iff & x^4 - 5x^2(x-1) + 5(x-1)^2 = 0\\
\iff & \left(x^2 - \frac52(x-1)\right)^2 = \left(\frac{25}{4} - 5\right)(x-1)^2 = \frac{5}{4}(x-1)^2\\
\iff & \left(x^2 - \frac{5 + \sqrt{5}}{2}(x-1)\right)\left(x^2 - \frac{5 - \sqrt{5}}{2}(x-1)\right) = 0
\end{align}
$$
