Is $f(x) = x^4 + x^2 + 1$ reducible in $\Bbb Z_2$?

Question

Is $f(x) = x^4 + x^2 + 1$ reducible in $\Bbb Z_2$?

My textbook says yes, and gives me the solution of $(x+1)^4$.. I don't quite see how to get there though. Before I checked the solution, I tried plugging in both $0$ and $1$ to see if I could pull out an $x - a$ term, but neither value gives a zero. Any tips?

Answer 1

Hint: A quartic can be reducible without having linear factors. Compute $(x^2+x+1)^2 \bmod 2$.

Answer 2

Like you've observed, $x^4 + x^2 + 1$ has no linear factors, so if it factors, it is a product of two irreducible quadratics.

An occasionally useful fact is that the only irreducible quadratic polynomial over $\Bbb Z_2$ is $x^2 + x + 1$ (proof: there are only three other quadratic polynomials; check them), so the only possibility in this case is $(x^2 + x + 1)^2$. Either this is or isn't $f(x)$.

Answer 3

Alt. hint:   try in $\,\mathbb{R}[\text{x}]\,$, first: $\,x^4+x^2+1\color{red}{+x^2-x^2}=(x^2+1)^2 - x^2 = \ldots\,$