<s>[INST] Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But this gives a polynomial of k where it has $\sqrt{k}$ terms as well. Is my approach correct or is there a simpler way? Please any hint would be highly appreciated.
 [/INST] Just do the Euclidean division of the two polynomials. You get
$$f_k(x)=\left(x^2-k\right)\left(2x^2+(3k-4)x+2k^2-3k-5\right)+(2k^3+k^2-7k-6)x+2k^3-3k^2-5k+6$$
And we want the remainder
$$(2k^3+k^2-7k-6)X+(2k^3-3k^2-5k+6)$$
to be the null polynomial. This is possible for the common roots if any of
$$\begin{align}2k^3+k^2-7k-6=&(2k+3)(k+1)(k-2)\\
2k^3-3k^2-5k+6=&(2k+3)(k-1)(k-2)\end{align}$$
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