A non-zero polynomial with real coefficients has the property that $f(x)=f'(x).f''(x)$.Then find the leading coefficient of $f(x).$ A non-zero polynomial with real coefficients has the property that $f(x)=f'(x).f''(x)$.Then find the leading coefficient of $f(x).$
I let $f(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+.....+a_n$ and then i found $f'(x)$ and $f''(x)$ and put in the given equation.But it becomes complicated involving many variables including $n$.Is my method wrong?Is there some other better method to solve it.
 A: You can first determine what $n$ is, since, you can first prove that, if $f$ is not $0$, then the degree of $f$ must be at least $2$. 
Now, if $n\geq 2$, then the degree of $f$ is $n$, the degree of $f'$ is $n-1$ and the degree of $f''$ is $n-2$. The degree of $f'\cdot f''$ is therefore $n-1 + n-2 = 2n-3$
Therefore, you want an integer that satisfies the equation $n = 2n-3$ or $n=3$.
A: You need to find the leading coefficient of $f(x)$, so assume $f(x)=ax^n$.
Then $f'(x)=anx^{n-1}$ and $f''(x)=an(n-1)x^{n-2}$.
Using the condition $f(x)=f'(x)\cdot f''(x)$, we have...
$$ax^n=[anx^{n-1}][an(n-1)x^{n-2}].$$
$$ax^n=a^2n^2(n-1)x^{2n-3}$$
Equating exponents, we have $n=2n-3$, which gives $n=3$.
Equating coefficients, we have...
$$a=a^23^2(3-1)$$
$$a=18a^2$$
$$a(18a-1)=0$$
Therefore, $a=0$, so $f(x)=0$, or $a=1/18$ which gives $\boxed{f(x)=\frac{1}{18}x^3}$.
A: Following my comment on Tim Thayer's solution,
Let $f(x)=\frac{x^3}{18}+ax^2+bx+c$.  Replace $x$ by $x+6a$ to remove the quadratic term, and now 
$$f(x)=\frac{x^3}{18}+dx+e\\
=f'(x)f''(x)=(\frac{x^2}6+d)(\frac x3)$$
So $d=d/3$; so $d$ and $e$ are both zero, and $f(x)=(x-m)^3$ are the only solutions.
