Finding the polynomial

Question

Find a nontrivial polynomial function $p(x)$ such that $p(2x)=p'(x)p''(x)\not=0$

Answer 1

If $\deg(p(x)) = n$, then we must have

\begin{align} deg(p'(x)p''(x)) &= deg(p(2x)) \\ (n-1) + (n-2) &= n \\ n &= 3 \end{align}

So let $p(x) = ax^3 + bx^2 + cx + d\;$ where $\;a \ne 0$:

\begin{align} p'(x) p''(x) = p(2x) \\ 18a^2 x^3 + 18ab x^2 + (6ac + 4b^2) x + 2bc &= 8a x^3 + 4b x^2 + 2c x + d \\ \hline 18a^2 &= 8a \\ 18ab &= 4b \\ 6ac + 4b^2 &= 2c \\ 2bc &= d \\ \hline a &= \dfrac 49 \\ b &= 0 \\ c &= 0 \\ d &= 0 \end{align}

So $p(x) = \dfrac 49 x^3$

Sorry I got it wrong the first time.

Answer 2

So you've got $P(x)=Ax^3+Bx^2+Cx+D$ from @Steven Gregory's post. Then $$\begin{align}8Ax^3+4Bx^2+2Cx+D&=(3Ax^2+2Bx+C)(6AX+2B)\\ &=18A^2x^3+18ABx^2+(4B^2+6AC)x+2BC\end{align}$$ If $A=0$, then we find that $P(x)=0$, so $A=\frac49$, but the $x^2$ term only works if $B=0$, then $C=D=0$ so the polynomial is just $\frac49x^3$.

Answer 3

Steven Gregory's answer may evolve further in the coming minutes. Let's try a more concrete approach: $$ f(x) = ax^3 + bx^2 + cx + d $$ (Third degree for reasons apparent from Steven Gregory's answer and the accompanying comments.) \begin{align} f(2x) & = 8ax^3 + 4bx^2 + 2cx + d \\ f'(x) & = 3ax^2 + 2bx + c \\ f''(x) & = 6ax + 2b \\ f'(x)f''(x) & = 18a^2 x^3 + 18abx^2 + (6ac+4b^2) x + 2bc \end{align} Therefore \begin{align} 8a & = 18 a^2 \\ 4b & = 18ab \\ 2c & = 6ac+4b^2 \\ d & = 2bc \end{align} \begin{align} a & = \frac 4 9 \\[10pt] b & = 0 \\[10pt] c & = 0 \\[10pt] d & = 0 \end{align}

If $f(x) = \dfrac 4 9 x^3$ then $f(2x) = \dfrac{32} 9 x^3$ and $f'(x) = \dfrac 4 3 x^2 $ and $f''(x) = \dfrac 8 3 x$, $\text{so } f'(x)f''(x) = \dfrac{32} 9 x^3.$

Answer 4

The degrees of each side must match: $\deg{p(2x)} = \deg\{p'(x) p''(x)\}$

Gives: $ n = (n-1) + (n-2)$ $ \implies n = 3$

Now we know the degree is 3, try setting $p(x)=a_3 x^3+a_2 x^2 + a_1 x^1+a_0$. Then evaluate $p(2x)$ and $p'(x) p''(x)$ and equate coefficients.