<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] I would begin by replacing $x^2$ with $k$ in the polynomial.  Where there is an odd power of $x$ you leave the remaining power of $x$ as is.  Thus
$2x^4=2k^2$
$(3k-4)x^3=(3k^2-4k)x$
$(2k^2-5k-5)x^2=2k^3-5k^2-5k$
$(2k^3-2k^2-3k-6)x=(2k^3-2k^2-3k-6)x$
$6=6$
Add the right sides together and collect all $x$ terms to get
$\color{blue}{(2k^3+k^2-7k-6)}x+\color{blue}{(2k^3-3k^2-5k+6)}$
If $x^2-k$ is to be a factor of the original polynomial, then the last expression above must be zero for *both roots** of $x^2=k$.  This is possible only if the blue polynomials are both zero for some common root $k$.  You are to find this common root.  It turns out that both blue polynomials have all their roots rational, so you should be able to find the answer with the usual method of searching for rational roots.
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