Question

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

Answer 1

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 `(x+1)`

Using synthetic division
`(x^5+x^3+x^2+1)div (x+1)`
yields the factors
`(x+1)(x^4-x^3+x-1)`

Again for `(x^4-x^3+x-1) = 0`
we have an obvious solution, `x=1`
So `(x-1)` is a factor

Using synthetic division again
`(x^4-x^3+x-1)div(x-1)`
gives the factors
`x^4-x^3+x-1) = (x-1)(x^3+1)`

Once again
`(x^3+1)=0`
has an obvious solution
`x=-1`
which leads to factors
`(x^3-1) = (x+1)(x^2-x+1)`

The complete factorization is
`x^5+x^3+x^2+1`

`= (x+1)^2(x-1)(x^2-x+1)`