Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ 
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$

....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial.....but it didnt help either.
 A: Write  $$x^4 - 12 x^2 + 8 x + 3=q(x)(x^2 + kx + 1)$$
then $q$ must be quadratic, so $q(x)= ax^2+bx+c$. By expanding and comparing the coeficients you will get $k$ (clearly $a=1$ and $c=3$)...  
A: We factor $x^4-12x^2+8x+3$.
We see, that $x=1$ is a root of $f(x)=x^4-12x^2+8x+3$, since $f(1)=1-12+8+3=0$
After long division $(x^4-12x^2+8x+3)\div (x-1)=x^3+x^2-11x-3$
We see, that x=3 is a root of the polynomial $g(x)=x^3+x^2-11x-3$
Again after long division we achieve:
$(x^3+x^2-11x-3)\div (x-3)=x^2+4x+1$
Which gives us the solution $k=4$.
A: Observe that there is no cubic term, and the independent term is $3$. Hence the factorization must be
$$(x^2+kx+1)(x^2-kx+3)=x^4+(4-k^2)x^2+2kx+3.$$
By identification $k=4$.

The "standard" solution is by equating the remainder of the long division,
$$(8+14k-k^3)x+16-k^2$$ to zero. The only solution is $k=4$.
A: Let $x$ be a root of the polynomial $x^2+kx+1$, thus $x^2=-(kx+1)$. By substitute it into ${x}^{4}-12\,{x}^{2}+8\,x+3$ we get  ${x}^{4}-12\,{x}^{2}+8\,x+3=\left( -{k}^{3}+14\,k+8 \right) x-{k}^{2}+16 \equiv 0$.  Then solve system 
\begin{cases}
 -{k}^{3}+14\,k+8=0,\\
-{k}^{2}+16=0,
\end{cases}
we get that $k=4.$
