# How do you factor x^5+x^3+x^2+1?

Apr 17, 2015

Since ${x}^{n}$ is negative for odd values of $n$
an obvious "solution" (if we pretend this expression $= 0$)
is $x = - 1$
Therefore one of the factors is $\left(x + 1\right)$

Using synthetic division
$\left({x}^{5} + {x}^{3} + {x}^{2} + 1\right) \div \left(x + 1\right)$
yields the factors
$\left(x + 1\right) \left({x}^{4} - {x}^{3} + x - 1\right)$

Again for $\left({x}^{4} - {x}^{3} + x - 1\right) = 0$
we have an obvious solution, $x = 1$
So $\left(x - 1\right)$ is a factor

Using synthetic division again
$\left({x}^{4} - {x}^{3} + x - 1\right) \div \left(x - 1\right)$
gives the factors
x^4-x^3+x-1) = (x-1)(x^3+1)

Once again
$\left({x}^{3} + 1\right) = 0$
has an obvious solution
$x = - 1$
$\left({x}^{3} - 1\right) = \left(x + 1\right) \left({x}^{2} - x + 1\right)$
${x}^{5} + {x}^{3} + {x}^{2} + 1$
$= {\left(x + 1\right)}^{2} \left(x - 1\right) \left({x}^{2} - x + 1\right)$