<s>[INST] Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are: 
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$ 
Is it right?
 [/INST] You can write it as $x^4+ax^3+bx^2-8x+4=(x^2+cx+d)^2$, which you want to be true for all $x$.  This requires that the coefficients of each power match.  You should expand the square on the right and match the coefficients.  This will give you each of $a,b,c,d$
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