# How do you factor x^5 - 5x^4 - x^3 + x^2 + 4 = 0?

May 5, 2015

Start by noticing that both $x = 1$ and $x = - 1$ are solutions, so $\left(x - 1\right)$ and $\left(x + 1\right)$ are both factors.

Then solve:
$\left(x - 1\right) \left(x + 1\right) \left(a {x}^{3} + b {x}^{2} + c x + d\right) = {x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4$
for $a$, $b$, $c$ and $d$:

Expanding:
$\left(x - 1\right) \left(x + 1\right) \left(a {x}^{3} + b {x}^{2} + c x + d\right) = \left({x}^{2} - 1\right) \left(a {x}^{3} + b {x}^{2} + c x + d\right)$
$= a {x}^{5} + b {x}^{4} + \left(c - a\right) {x}^{3} + \left(d - b\right) {x}^{2} - c x - d$

Comparing the coefficients of ${x}^{5}$, ${x}^{4}$, etc., we can quickly see that $a = 1$, $b = - 5$, $c = 0$ and $d = - 4$.

So, thus far we have:
${x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4 = \left(x - 1\right) \left(x + 1\right) \left({x}^{3} - 5 {x}^{2} - 4\right)$.

Differentiating the cubic ${x}^{3} - 5 {x}^{2} - 4$ gives $3 {x}^{2} - 10 x$. This is zero for $x = 0$ and $x = \frac{10}{3}$. These are the $x$ coordinates of the turning points of the cubic. Substituting each of these two values of $x$ into the cubic give negative values, so the cubic has only 1 real root - the other two are complex numbers. Notice that ${x}^{3} - 5 {x}^{2} - 4$ is negative for $x = 5$ and positive for $x = 6$. So the real root lies somewhere between 5 and 6, but it is not a rational number, let alone an integer. So the factorisation above is 'complete'.