Find k so that polynomial x2+2x+k …

If g(x) = x² + 2x + k is a factor of  f(x) = 2x⁴ + x³ - 14x² + 5x + 6, then remainder is zero when f(x) is divided by g(x).
Let quotient = Q and remainder = R
Let us now divide f(x) by g(x).

R = x(7k + 21) + (2k² + 8k + 6) -------(1) and  Q = 2x² - 3x - 2(k + 4).------------(2)
Now, R = 0.
{tex}\Rightarrow{/tex} x (7k + 21) + 2 (k² + 4k + 3) = 0 
{tex}\Rightarrow{/tex} 7x (k + 3) + 2 (k+1)(k+3) = 0
{tex}\Rightarrow{/tex} (k+3) [7x + 2(k+1)] = 0
{tex}\Rightarrow{/tex} k + 3 = 0
{tex}\Rightarrow{/tex} k = -3
Thus, polynomial f(x) can be written as,
2x⁴ + x³ - 14x² + 5x + 6 = (x² + 2x + k) [2x² - 3x - 2(k + 4)] = (x² + 2x - 3) (2x² - 3x - 2)

Zeros of x² + 2x - 3 are,
x² + 2x - 3 = 0
{tex}\Rightarrow{/tex} (x + 3) (x - 1) = 0
{tex}\Rightarrow{/tex} x = -3 or x = 1

Zeros of (2x² - 3x - 2) are,
2x² - 3x - 2 = 0
{tex}\Rightarrow{/tex} 2x² - 4x + x - 2 = 0
{tex}\Rightarrow{/tex} 2x(x - 2) + 1(x - 2) = 0
{tex}\Rightarrow{/tex} (x - 2)(2x + 1) = 0
x = 2 or x = -{tex}\frac12{/tex}
Thus, the zeros of f(x) are: -3 ,1, 2 and -{tex}\frac12{/tex}