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:

 A: 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 $x^2-ax+1$.
A: Suppose that you have a monic (e.g. leading coefficient of 1) 4th degree polynomial $x^4 + ax^3 + bx^2 + cx + 1$ that you factor into two 2nd degree polynomials:
$(ex^2 + fx + g) \times (hx^2 + ix + j).$
Then, you can divide each coefficient of the first polynomial by $e$ and multiply each coefficient of the second polynomial by $e$.  This produces:
$(x^2 + [f/e]x + [g/e]) \times ([he]x^2 + [ie]x + [je]).$
However, since the product of these two polynomials is 
$x^4 + ax^3 + bx^2 + cx + 1$, 
then $h \times e$ must = 1.$
Therefore, the 4th degree monic polynomial has been factored into two 2nd degree monic polynomials.  As others have pointed out, under this factoring, just because the $x^0$ coefficient in the 4th degree polynomial is 1 doesn't mean that the $x^0$ coefficients in the two 2nd degree polynomials each have to be one.  All you can say for sure is that the product of the two $x^0$ coefficients in the two 2nd degree polynomials must = 1.
If I understand correctly, it just so happened that when the monic 4th degree polynomial given in the original query is factored into two monic 2nd degree coefficients, for that particular 4th degree coefficient, the resulting monic 2nd degree polynomials happen to have their $x^0$ coefficients each = 1.
Addendum Focusing on the OP's original 4th degree polynomial
First of all, consider the 4th degree polynomial that equals 
$(x^2 + x + 5) \times (x^2 + x + [1/5]).$ 
This is a simple counter-example whose product will have form
$x^4 + ax^3 + bx^2 + cx + 1.$
Edit
Well, this is embarasssing:
I just realized that my counter-example above is flawed.  That is, when
$(x^2 + x + 5) \times (x^2 + x + [1/5])$ is combined into a monic 4th degree polynomial, there may well be alternative ways of factoring this 4th degree polynomial that fit the pattern that was originally suggested to the OP.
Anyway, the remainder of this addendum looks at the constraints in a manner very similar to the What is the meaning of symmetry of the coefficients? link that someone already commented.
All of this analysis begs the question of why there apparently was a
suggestion to factor
$f(x)  =  x^4 - x^3 + x^2 - x + 1$ into 
$(x^2 - ax + 1) \times (x^2 - bx + 1).$
I surmise that what is really happening is that it has been
conjectured that $f(x)$ can be so factored.
Consequently, the student is being asked to explore the conjecture,
and see if it is true.  Exploring leads to the following constraints on
$a$ and $b$:
(1) re $x^3 : a + b = 1.$ 
(2) re $x^2 : 2 + (a \times b) = 1.$ 
(3) re $x^1 : a + b = 1.$
Notice that you have three constraints on the two variables
$a$ and $b.$
However, since constraints (1) and (3) happen to be identical, you
end up with only two constraints.
Even if both constraints (1) and (2) were linear, this still wouldn't
(in general) guarantee a solution [e.g. r + s = 6. 2r + 2s = 11].
In the present case, constraint (2) is non-linear, which makes it
even more iffy.  Note: I am on thin ice here, I've never studied the
effect of combining 1 linear constraint with 1 non-linear constraint.
However, exploring as intended, presumably, satisfying values of
$a$ and $b$ can be found.  Taking a look at $f(x),$ notice that
constraint (3) is identical to constraint (1) precisely because in
$f(x)$ the $x^3$ and $x^1$ coefficients are identical.
Therefore, it could be argued that the suggested conjecture was
well motivated.
