Find k so that x2 + …

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 gx)

R = x(7k + 21) + (2k² + 8k + 6) -------(1)

and  Q = 2x² - 3x - 2(k + 4).------------(2)
{tex}\Rightarrow{/tex}x (7k + 21) + 2 (k² + 4k + 3) = 0 
{tex}\Rightarrow{/tex}7k + 21 = 0 and k² + 4k + 3 = 0
{tex}\Rightarrow{/tex} 7(k + 3) = 0 and (k + 1) (k + 3) = 0
{tex}\Rightarrow{/tex} k + 3 = 0
{tex}\Rightarrow{/tex}k = -3
Substituting the value of k in the divider x² + 2x + k, we obtain: x² + 2x - 3 = (x + 3) (x - 1) as the divisor.
Hence two zeros of g(x) are -3 and 1.------(3)
Putting k=-3 in (2) we get
Q = 2x² - 3x - 2
= 2x² - 4x + x - 2
= 2x(x - 2) + 1 (x - 2)
= (x - 2)(2x + 1)

Q=0 if x-2=0 or 2x+1=0
So other two zeros of f(x) are 2 and -{tex}\frac12{/tex}-------(4)

As g(x) is a factor of f(x) so zeros of G(x) are zeros of f(x) also

Hence from (3) and (4) we get
The zeros of f(x) are: -3 ,1,, 2 and  {tex}\frac12{/tex}