Polynomial of 4th degree I would like to ask if someone could help me with the following equation.
\begin{equation}
x^4+ax^3+(a+b)x^2+2bx+b=0
\end{equation}
Could you first solve in general then $a=11$ and $b=28$.
I get it to this form but I stuck.
\begin{equation}
1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0
\end{equation}
Thank you in advance.

The polynomial equation for $(a, b)=(11,28)$ is given by
$$
x^4+11x^3+39x^2+56x+28=(x^2 + 7x + 7)(x + 2)^2=0.
$$
"How do we come up wit this layout"? By the rational root test, we find the factor $x-2$ twice, and dividing gives the factor $x^2+7x+7$. Hence the roots are $x=-2,-2,\frac{\sqrt{21}-7}{2},\frac{-\sqrt{21}-7}{2}$.
In general, the polynomial will not have any integral roots. A good example is the case $(a, b)=(1,1)$, where the polynomial
$$
x^4+x^3+2x^2+2x+1
$$
is irreducible over $\Bbb Q$. It has no real root at all.
