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Suppose we are given the fact that if $g(x)$ is irreducible and $g(x)$ divides both $f(x)$ and $f'(x)$, then $(g(x))^2$ divides $f(x)$. How do we factor $x^6+x^4+3x^2+2x+2$ into a product of irreducible over the complex numbers?

Attempt: I have tried the rational roots and there weren't any. I differentiated and integrating $f$ but a fifth and seventh degree polynomial wasn't any easier to work with as far as I can tell. I tried brute force long division by dividing some arbitrary $x^2+Ax+B$, and it was a mess. Currently I'm out of ideas.

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    The information given -- namely, a result about polynomials dividing both $f$ and $f'$ -- suggests that maybe you should try and find polynomials dividing both your $6$th degree polynomial, and its derivative (i.e., compute their GCD). Have you tried that? – 2017-01-16
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    The obvious way to go has been pointed out by @pjs36 already. That said, given the hint and if you trust your lucky stars, notice that there don't appear to be any "*nice*" linear factors, so try a factorization in the form $(x^2+ax+b)^2(x^2+cx+d)$. – 2017-01-16
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    @pjs36 ah I did not notice that I should find the GCD thank you so much! – 2017-01-16
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    $(x^2-2x +2)( x^2 + x + 1)^2$ – 2017-01-16

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