# How do you factor y^5 - y^4 + y^3 - y^2 + y - 1?

Jun 8, 2015

$f \left(y\right) = {y}^{5} - {y}^{4} + {y}^{3} - {y}^{2} + y - 1$

Then $f \left(1\right) = 1 - 1 + 1 - 1 + 1 - 1 = 0$

So $\left(y - 1\right)$ is a factor.

$f \left(y\right) = \left(y - 1\right) \left({y}^{4} + {y}^{2} + 1\right)$

${y}^{4} + {y}^{2} + 1$ has no linear factors since ${y}^{4} + {y}^{2} + 1 \ge 1 > 0$ for all $y \in \mathbb{R}$

Suppose ${y}^{4} + {y}^{2} + 1 = \left({y}^{2} + a y + b\right) \left({y}^{2} + c y + d\right)$ for some constants
$a$, $b$, $c$ and $d$.

$\left({y}^{2} + a y + b\right) \left({y}^{2} + c y + d\right)$

$= {y}^{4} + \left(a + c\right) {y}^{3} + \left(d + a c + b\right) {y}^{2} + \left(a d + b c\right) y + b d$

Comparing coefficients, we get:

[1] $a + c = 0$

[2] $d + a c + b = 1$

[3] $a d + b c = 0$

[4] $b d = 1$

From [1] we get $c = - a$, so we can rewrite:

[2a] $b + d = {a}^{2} + 1$

[3a] $a \left(d - b\right) = 0$

From [3a] we can deduce that $a = 0$ or $b = d$.

Suppose $a = 0$. Then [2a] gives us $b + d = 1$. Together with [4], this gives us $b + \frac{1}{b} = 1$, hence ${b}^{2} - b + 1 = 0$ which has no real solutions since its discriminant is $- 3$,

The other option is $b = d$. Together with [4], this means $b = d = 1$ or $b = d = - 1$. From [2a] we know $b + d \ge 1$ so we can discount the possibility $b = d = - 1$ and focus on $b = d = 1$.

From [2a] we get ${a}^{2} = 2 - 1 = 1$, so $a = \pm 1$

In a rather lengthy way, we have found:

${y}^{4} + {y}^{2} + 1 = \left({y}^{2} + y + 1\right) \left({y}^{2} - y + 1\right)$

So

$f \left(y\right) = \left(y - 1\right) \left({y}^{2} + y + 1\right) \left({y}^{2} - y + 1\right)$