TITLE: Decomposition of polynomial into irreducible polynomials
QUESTION [7 upvotes]: I'm preparing to my algebra exam. And I have problem and I have no idea how to solve it.

Given polynomial
  $$x^4+4x^3+4x^2+1.$$
  The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$.

I will be happy if you show me the way how to solve such problems

REPLY [6 votes]: It suffices to show that $f(x)=x^4+4x^3+4x^2+1$ has no linear factors over $\mathbb{Z}_3$: $f(0)=f(1)=1$ and $f(2)=2$, so $f(x)$ has no linear factors.  Then $f(x)$ must factor to two quadratic polynomials: $$f(x)=(ax^2+bx+c)(ux^2+vx+w)$$ We then have that $au=1$.  Multiplying the first polynomial by $u$ and the second by $a$, we may assume that $a=u=1$.  Equating the coefficients of the powers of $x$, we have
$$
\begin{eqnarray*}
4&=&v+b\\
4&=&w+c+bv\\
0&=&bw+cv\\
1&=&cw
\end{eqnarray*}
$$
Some algebra shows that $$\begin{eqnarray*}
 b&=& 2+\sqrt{2 \left(1+\sqrt{2}\right)}\\c&=& 1+\sqrt{2}+\sqrt{2 \left(1+\sqrt{2}\right)} \\ v&=& 2-\sqrt{2 \left(1+\sqrt{2}\right)}\\ w&=& 1+\sqrt{2}-\sqrt{2 \left(1+\sqrt{2}\right)} 
\end{eqnarray*}
$$
So, letting $\alpha=1+\sqrt{2}$, we see that $x^4+4x^3+4x^2+1$ factors to $$\left(x^2+\left(2+\sqrt{2\alpha}\right)x+\alpha+\sqrt{2\alpha}\right)\left(x^2+\left(2-\sqrt{2\alpha}\right)x+\alpha-\sqrt{2\alpha}\right).$$