<s>[INST] Problem with factoring $x^4-x^3+x^2-x+1$ I want to calculate following integral Using partial fraction: $$\int{1\over x^5+1}dx$$So I decompose the denominator:
$$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$$
For the next step I searched on internet and find out I should decompose$x^4-x^3+x^2-x+1$ like this:
$$x^4-x^3+x^2-x+1=(x^2-ax+1)(x^2-bx+1)$$
And then $a,b$ can be found easily.
My question is Why the coefficients of $x^2,x^0$ are $1$?
Because I can rewrite:
$$x^4-x^3+x^2-x+1=(ax^2+bx+c)(dx^2+ex+f)$$
And only thing I can see in the first look is $ad=1,cf=1$ and I have no clue that why $a=d=c=f=1$
You can see his answer below:

 [/INST] In general, the two polynomials are given up to multiplication of a constant (you can multiply one by $k$ and other by $1/k$), so you can arrange it in a way that $a=d=1$ is guaranteed. For example $x^2+4x+4$ can be factored as $(x+2)(x+2)$ but also as $(2x+4)(\frac{1}{2}x+1)$. So we are free to fix one of the coefficient to make the answer unique. However if you do this, then you don't have the choice for others, so a correct start here is something like
$$x^4-x^3+x^2-x+1=(x^2−ax+b)(x^2−cx+d).$$
Sure you can do some calculation further to get more information about the constant coefficients, but not before that.
Also following slightly modified example shows that assuming both leading and constant coefficients to be $1$ from the start is wrong:
$$
x^4-x^3+x^2+x+1=\\(x^2+0.86676039+0.46431261)(x^2-1.86676039x+2.15372137)
$$
However, as pointed in the linked other question, in this case it was probably used (but not explained) that the polynomial is palindromic (self-reciprocal), which implies its roots come in pairs $\alpha, \frac{1}{\alpha}$ (it's a result of $x^4f(1/x)=f(x)$). This allows you to expect the factors in a form $$(x-\alpha)(x-\frac{1}{\alpha})=x^2-(\alpha+\frac{1}{\alpha})x+1,$$ or more generic $x^2-ax+1$.
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