C-02-Q132: If $f(y) = y^3 + ky^2 - 4y - 8$ and $f(-2) = 0$, then $k =$

C-02-Q132: If $f(y) = y^3 + ky^2 - 4y - 8$ and $f(-2) = 0$, then $k =$ | প্রশ্ন করি

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If f(y)=y3+ky2−4y−8f(y) = y^3 + ky^2 - 4y - 8f(y)=y3+ky2−4y−8 and f(−2)=0f(-2) = 0f(−2)=0, then k=k =k=