C-02-Q132mediumsingle_mcqIf 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=a−2-2−2b000c222dব্যাখ্যাPutting y=−2y=-2y=−2: (−2)3+k(−2)2−4(−2)−8=−8+4k+8−8=4k−8=0(-2)^3+k(-2)^2-4(-2)-8=-8+4k+8-8=4k-8=0(−2)3+, which gives .