Abstract Algebra – Decomposition of Polynomial into Irreducible Polynomials

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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

Best Answer

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 $$\left(x^2+\left(2+\sqrt{2\alpha}\right)x+\alpha+\sqrt{2\alpha}\right)\left(x^2+\left(2-\sqrt{2\alpha}\right)x+\alpha-\sqrt{2\alpha}\right).$$

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