# Given polynomial x^4+4x^3+4x^2+1 The task is find expansion of the polynomial as a product of irreducible polynomials in RR

Given polynomial
${x}^{4}+4{x}^{3}+4{x}^{2}+1.$
The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$
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Helena Howard
It suffices to show that $f\left(x\right)={x}^{4}+4{x}^{3}+4{x}^{2}+1$ has no linear factors over ${\mathbb{Z}}_{3}$:
$f\left(0\right)=f\left(1\right)=1$ and $f\left(2\right)=2$, so $f\left(x\right)$ has no linear factors. Then $f\left(x\right)$ must factor to two quadratic polynomials:
$f\left(x\right)=\left(a{x}^{2}+bx+c\right)\left(u{x}^{2}+vx+w\right)$
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
$\begin{array}{rcl}4& =& v+b\\ 4& =& w+c+bv\\ 0& =& bw+cv\\ 1& =& cw\end{array}$
Some algebra shows that
$\begin{array}{rcl}b& =& 2+\sqrt{2\left(1+\sqrt{2}\right)}\\ c& =& 1+\sqrt{2}+\sqrt{2\left(1+\sqrt{2}\right)}\\ v& =& 2-\sqrt{2\left(1+\sqrt{2}\right)}\\ w& =& 1+\sqrt{2}-\sqrt{2\left(1+\sqrt{2}\right)}\end{array}$
So, letting $\alpha =1+\sqrt{2}$, we see that ${x}^{4}+4{x}^{3}+4{x}^{2}+1$ factors to
$\left({x}^{2}+\left(2+\sqrt{2\alpha }\right)x+\alpha +\sqrt{2\alpha }\right)\left({x}^{2}+\left(2-\sqrt{2\alpha }\right)x+\alpha -\sqrt{2\alpha }\right).$