<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] Factor out $x^2-k$:
$$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6=\\
[2x^4-2x^2k+2x^2k-2k^2+\color{red}{2k^2}]+\\
[(3k-4)x^3-(3k-4)xk+\color{red}{(3k-4)xk}]+\\
[(2k^2-5k-5)x^2-(2k^2-5k-5)k+\color{red}{(2k^2-5k-5)k}]+\\
\color{red}{(2k^3-2k^2-3k-6)x+6}\Rightarrow \\
\begin{cases}2k^3-2k^2-3k-6+(3k-4)k=0\\
2k^2+(2k^2-5k-5)k+6=0 \end{cases} \Rightarrow \\
\begin{cases}2k^3+k^2-7k-6=0\\
2k^3-3k^2-5k+6=0\end{cases} \Rightarrow \\
4k^2-2k-12=0 \Rightarrow \\
2k^2-k-6=0 \Rightarrow \\
k_{1,2}=\frac{1\pm 7}{4}=-\frac32,2.$$
Verification: WA answer.
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