<s>[INST] Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.
 [/INST] First you seem to know that $(x+1)|(x^5+1)$.  Carry out this division with polynomial long division to get
$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$
The remaining fourth degree polynomial is symmetric or palindromic, its coefficients $1,-1,1,-1,1$ read the same forward and backwards.  When this happens there will be quadratic factors that are palindromic too, so you must have
$x^4-x^3+x^2-x+1=(x^2+ax+1)(x^2+bx+1)$
Expand:
$x^4-x^3+x^2-x+1=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$
Matching like powers of $x$ gives $b=-1-a$ and $ab=-1$.  Eliminating $b$ between these equations gives a quadratic equation for $a$ which you can solve.
I leave you to prove:  (1) Both roots for $a$ give the same factors for $x^4-x^3+x^2-x+1$, only the order is different.  (2) The quadratic factors you find have negative discriminants and thus cannot be broken down to real linear factors.
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