Compute the quotient and remainder,
[tex]\dfrac{2x^4+x^3-14x^2+5x+6}{x^2+2x+k} \\\\ = 2x^2 - 3x - (8+2k) + \dfrac{(21+7k)x+(6+8k+2k^2)}{x^2+2x+k}[/tex]
The remainder upon dividing [tex]2x^4+x^3-14x^2+5x+6[/tex] by [tex]x^2+2x+k[/tex] should leave no remainder, which means
[tex]21+7k = 0 \implies 21 = -7k \implies k=-3[/tex]
and
[tex]6+8k+2k^2 = 0 \implies 2(k+3)(k+1)=0 \implies k=-3\text{ or }k=-1[/tex]
Only k = -3 makes both remainder terms vanish.
Then the previous result reduces to
[tex]\dfrac{2x^4+x^3-14x^2+5x+6}{x^2+2x-3} = 2x^2 - 3x - 2[/tex]
so that
[tex]2x^4+x^3-14x^2+5x+6 = (x^2+2x-3) (2x^2 - 3x - 2) \\\\ 2x^4+x^3-14x^2+5x+6 = (x+3)(x-1)(2x + 1)(x-2)[/tex]
and so the zeroes of the quartic polynomial are x = -3, x = 1, x = -1/2, and x = 2.
