The correct option is D 118
Let degree of f(x) be n.
Then, degree of f′(x)=n−1
and degree of f′′(x)=n−2
Hence n=(n−1)+(n−2)
⇒n=2n−3
⇒n=3
Now, let f(x)=ax3+bx2+cx+d, where a≠0
f′(x)=3ax2+2bx+c
f′′(x)=6ax+2b
∴ax3+bx2+cx+d=(3ax2+2bx+c)(6ax+2b)
Comparing the coefficient of x3,
18a2=a
⇒a=118