If g(x)=x2 + 2x + k is a factor of f(x) = 2×4 + x3 – 14×2 + 5x + 6Then remainder is zero when f(x) is divided by g(x)Let quotient =Q and remainder =RLet us now divide f(x) by gx)R = x(7k + 21) + (2k2 + 8k + 6) ——-(1)and Q = 2×2 – 3x – 2(k + 4).————(2){tex}\\Rightarrow{/tex}x (7k + 21) + 2 (k2 + 4k + 3) = 0\xa0{tex}\\Rightarrow{/tex}7k + 21 = 0 and k2 + 4k + 3 = 0{tex}\\Rightarrow{/tex}\xa07(k + 3) = 0 and (k + 1) (k + 3) = 0{tex}\\Rightarrow{/tex}\xa0k + 3 = 0{tex}\\Rightarrow{/tex}k = -3Substituting 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 getQ = 2×2 – 3x – 2= 2×2 – 4x + x – 2= 2x(x – 2) + 1 (x – 2)= (x – 2)(2x + 1)Q=0 if x-2=0 or 2x+1=0So 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) alsoHence from (3) and (4) we getThe zeros of f(x) are: -3 ,1,, 2 and {tex}\\frac12{/tex}