Polynomial satisfying $f(x) = f'(x) \cdot f''(x)$ If a polynomial of degree $n$ satisfies $f(x) = f'(x)\cdot f''(x)$ (such that $n$ belongs to $\mathbb R$) then $f(x)$ is?
A) an onto function 
B) an into function 
C) no such function possible 
D) even function 
I tried this question by letting a polynomial $f(x) = ax^n + bx^{n-1} \cdots$ and then derivated it but it did not help much , then i assumed a quadratic expression $ax^2 + bx + c$ and got $ax^2 + x(b-4a^2)$. 

Suppose that $n\geq2$. Then we have that the degree of the left side is $n$, but that of the right side is $(n-1)+(n-2)=2n-3$.
So we want to solve $2n-3=n$, or $n=3$. 
Then write $f(x)=ax^3+bx^2+cx+d$ with $a \neq 0$, then we have $f'(x)= 3ax^2+2bx+c$ and $f''(x)= 6ax+2b$. Thus $$f'(x)f''(x) = (3ax^2+2bx+c)(6ax+2b) = 18a^2x^3+18abx^2+(6ac+4b^2)x+2bc$$
Now we have $18a^2=a$, thus $a = \frac{1}{18}$. The second coefficient doens't give us anything. The third gives $\frac{1}{3}c+4b^2=c$, and thus $b^2=\frac{1}{6}c$. Furthermore we have $2bc=d$, so $12b^3=d$. This gives the following equation for $f(x)$:
$$f(x)=\frac{1}{18}x^3 + bx^2+ 6b^2x+12b^3$$
for some $b \in \mathbb R$.
If $n$ is 1, the LHS is a linear function and the RHS a constant.
If $f$ is a constant unequal to zero, the RHS is a constant equal to zero.
This leaves only the zero polynomial $f(x)=0$, that indeed statistifies the equation. 

Conclusion:
The only polynomials that statistify the given equations are $f(x)=0$ and 
$$f(x)=\frac{1}{18}x^3 + bx^2+ 6b^2x+12b^3$$
for some $b \in \mathbb R$.
