If $p(x)$ be a polynomial satisfying $p(2 x)=p^{\prime}(x) \cdot p^{\prime \prime}(x)$, then $\sum_{x=1}^5 p(x)=$

It is given a polynomial satisfying $p(2 x)=p^{\prime}(x) p^{\prime \prime}(x)$ Let the degree of polynomial is $n$, then degree of $p^{\prime}(x)$ and $p^{\prime \prime}(x)$ are $n-1$ and $n-2$ respectively, so $n=n-1+n-2 \Rightarrow n=3$ Now, let the polynomial $p(x)=a x^3$ $\therefore \quad p(2 x)=p^{\prime}(x) \cdot p^{\prime \prime}(x) \Rightarrow 8 a x^3=\left(3 a x^2\right) \cdot(6 a x)$ $\Rightarrow \quad a=\frac{4}{9}$ $\therefore$ The required polynomial is $p(x) \frac{4}{9} x^3$ Therefore $\sum_{x=1}^5 p(x)=\frac{4}{9}\left[1^3+2^3+3^3+4^3+5^3\right]$ $=\frac{4}{9}\left(\frac{5 \times 6}{2}\right)^2=100$