<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] Use that $$\frac{f(x)}{x^2-k}=2\,{k}^{2}+3\,kx+2\,{x}^{2}-3\,k-4\,x-5+{\frac {2\,{k}^{3}x+2\,{k}^{3}
+{k}^{2}x-3\,{k}^{2}-7\,kx-5\,k-6\,x+6}{{x}^{2}-k}}
$$
</s>