MHB Can You Solve This Factor Problem with $x^5+x^4+1$?

[kaliprasad]

factor $x^5+x^4+1$

 

[mathbalarka]

Spoiler

$$\begin{align} x^5 + x^4 + 1 = x^5 + x^4 + x^3 + x^2 + x + 1 - x^3 - x^2 - x & = x^3(x^2 + x + 1) + 1(x^2 + x + 1) - x (x^2 + x + 1) \\ &= (x^3 - x + 1)(x^2 + x + 1)\end{align}$$

EDIT : In particular, $x^5 + x^4 + 1 = 0 \Rightarrow t^5 + t + 1 = 0$ where $t = 1/x$. $t^n + t + 1$ has always a factor of the form $t^2 + t + 1$ for $t = 2 \bmod 3$.

 

[Dethrone]

No elegance to this approach, but:

It has no linear factors, so they must factor as:
$$(x^2+ax+b)(x^3+cx^2+dx+e)$$

Solving the resulting system of equations:
$a=1$, $b=1$, $c=0$, $d=-1$, $e=1$

Therefore,
$$x^5+x^4+1=(x^2+x+1)(x^3-x+1)$$

 

[mathbalarka]

Spoiler

Another solution, maybe?

$$x^5 + x^4 + 1 = x^5 + x^4 - x^2 + 1 + x^2 = x^5 - x^2 + x^4 + x^2 + 1 = x^2(x^3 - 1) + (x^4 + x^2 + 1) = x^2(x - 1)(x^2 + x + 1) + (x^2 + x + 1)(x^2 - x + 1) = (x^2 + x + 1)(x^3 - x + 1)$$

 

[kaliprasad]

Spoiler

By rational root theorem it does not have a rational root and so it shall be product of a quadratic function and a cubic function

as it is of the form $x^{3n+2} + x^{3m+1} + 1$ so
$x = \omega$ and $x = \omega^2$ are zeros ($\omega$ is cube root of 1)

$(x-\omega)(x-\omega^2)$ or $x^2 + x + 1$ is a factor and by division

$x^5 + x^4 + 1 = (x^2+x+1)(x^3 - x + 1)$