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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.

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    $\begingroup$ Do you mean $f(x)=f'(x) \cdot f''(x)$? $\endgroup$
    – Asydot
    Sep 26, 2015 at 16:00
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    $\begingroup$ What is the leading power $x^k$ for $f'(x)f''(x)$? It must be the same degree as for $f(x)$ $\endgroup$
    – Empy2
    Sep 26, 2015 at 16:00

3 Answers 3

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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 $n=3$.

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    $\begingroup$ +1: I think you meant to say "if $f$ is not $0$, then the degree of $f$ must be at least $2$," no? $\endgroup$ Sep 26, 2015 at 16:09
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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 $\boxed{f(x)=\frac{1}{18}x^3}$.

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  • $\begingroup$ You also have $f(x)=(x-a)^3/18$. OP didn't ask, but are they the only solutions? $\endgroup$
    – Empy2
    Sep 26, 2015 at 16:17
  • $\begingroup$ @Michael I don't understand your comment. Multiply out $(x-a)^3$ and the leading term is $x^3$, which is what I started with. From the work, my solution is unique. $\endgroup$
    – Tim Thayer
    Sep 26, 2015 at 16:20
  • $\begingroup$ Sorry, your solution is correct. But my polynomials also satisfy $f(x)=f'(x)f''(x)$. They have the same leading coefficient, which you proved, but can there be different lower coefficients? $\endgroup$
    – Empy2
    Sep 26, 2015 at 16:22
  • $\begingroup$ What are your polynomials? Certainly an answer like $x^3/18+x^2-2x$ would also satisfy, but the question asks for the leading coefficient, so only the leading term matters. $\endgroup$
    – Tim Thayer
    Sep 26, 2015 at 16:24
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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.

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