Find $k$ such that $x^{2}+2x+k$ is a factor of $2x^{4}+x^{3}-14x^{2}+5x+6$. Also,find all the zeroes of the two polynomials.

(C) Given that $(x^{2}+2x+k)$ is a factor of $2x^{4}+x^{3}-14x^{2}+5x+6$,the remainder must be zero when we perform polynomial division.
Performing long division of $2x^{4}+x^{3}-14x^{2}+5x+6$ by $(x^{2}+2x+k)$:
$1$. Divide $2x^{4}$ by $x^{2}$ to get $2x^{2}$. Multiply $(x^{2}+2x+k)$ by $2x^{2}$ to get $2x^{4}+4x^{3}+2kx^{2}$. Subtracting this from the dividend gives $-3x^{3}-(2k+14)x^{2}+5x+6$.
$2$. Divide $-3x^{3}$ by $x^{2}$ to get $-3x$. Multiply $(x^{2}+2x+k)$ by $-3x$ to get $-3x^{3}-6x^{2}-3kx$. Subtracting this gives $(6-2k-14)x^{2}+(3k+5)x+6 = (-8-2k)x^{2}+(3k+5)x+6$.
$3$. Divide $(-8-2k)x^{2}$ by $x^{2}$ to get $(-8-2k)$. Multiply $(x^{2}+2x+k)$ by $(-8-2k)$ to get $(-8-2k)x^{2} + 2(-8-2k)x + k(-8-2k)$.
Subtracting this from the previous remainder gives the final remainder: $(3k+5+16+4k)x + (6+8k+2k^{2}) = (7k+21)x + (2k^{2}+8k+6)$.
For the remainder to be zero,both coefficients must be zero:
$7k+21=0 \Rightarrow k=-3$.
$2k^{2}+8k+6=0 \Rightarrow 2(k^{2}+4k+3)=0 \Rightarrow 2(k+1)(k+3)=0 \Rightarrow k=-1$ or $k=-3$.
Since $k$ must satisfy both,$k=-3$.
With $k=-3$,the divisor is $x^{2}+2x-3 = (x+3)(x-1)$,so its zeroes are $x=-3, 1$.
The quotient is $2x^{2}-3x-2 = 2x^{2}-4x+x-2 = 2x(x-2)+1(x-2) = (2x+1)(x-2)$,so its zeroes are $x=-1/2, 2$.
Thus,the zeroes of $x^{2}+2x-3$ are $1, -3$ and the zeroes of $2x^{4}+x^{3}-14x^{2}+5x+6$ are $1, -3, 2, -1/2$.