Do the grouping $k^{4}-4k^{3}-k+4=\left(k^{4}-4k^{3}\right)+\left(-k+4\right)$, and factor out $k^{3}$ in the first and $-1$ in the second group.
$$k^{3}\left(k-4\right)-\left(k-4\right)$$
Factor out common term $k-4$ by using distributive property.
$$\left(k-4\right)\left(k^{3}-1\right)$$
Consider $k^{3}-1$. Rewrite $k^{3}-1$ as $k^{3}-1^{3}$. The difference of cubes can be factored using the rule: $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$.
$$\left(k-1\right)\left(k^{2}+k+1\right)$$
Rewrite the complete factored expression. Polynomial $k^{2}+k+1$ is not factored since it does not have any rational roots.
