Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$

Question

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.

Answer 1

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

Answer 2

Your approach is good. Just add $f(\sqrt k)+f(-\sqrt k)=0$ to get rid of the terms with $\sqrt k$.

Answer 3

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.

Answer 4

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.

Answer 5

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

$$\begin{align}2k^3+k^2-7k-6=&(2k+3)(k+1)(k-2)\\ 2k^3-3k^2-5k+6=&(2k+3)(k-1)(k-2)\end{align}$$