Decomposition of polynomial into irreducible polynomials

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

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@FlybyNight - you get a real factorisation into quadratics by organising the roots into conjugate pairs. – Mark Bennet Jan 12 '13 at 15:46
@Fly: $\mathbb C$ is a quadratic extension, so every polynomial over $\mathbb R$ of degree greater than two is reducible. The only ireducible polynomials are linear ones and quadratics with negative determinant. – Marek Jan 12 '13 at 15:48

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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Of course, a system! I would say that the ideal answer would be a combination of Mark Bennet's and Marek's comment to estabilish that the asked factorization exists, combined with the system of equations. – Git Gud Jan 12 '13 at 16:15

Hint:$$x^4+4x^3+4x^2+1=(x+1)^4-2(x+1)^2+2$$ $$(x+1)^4-2(x+1)^2+2=((x+1)^2-1)^2+1=((x+1)^2-1+i)((x+1)^2-1-i)$$ This will allow you to get the roots of the equation $x^4+4x^3+4x^2+1=0$. By multiplying the monomials with conjugate roots you will get the real quadratic factors.

To see why this always works:

It is a fact that if $P$ is a polynomial such that $P(z)=0$, then $P(\bar z)=0$.

Finally:

$$(x-(a+bi))(x-(a-bi))=x^2-2ax+a^2+b^2$$

Which is a real quadratic polynomial.

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How do you get the first equality just like that? – Git Gud Jan 12 '13 at 16:06
I used an online software that gets all of the roots of a polynomial. After seeing that all of them were in the form -1+... I tried substituting $x=u-1$ and it worked. – Amr Jan 12 '13 at 16:08