If p(x) is a polynomial satisfying p(2x) = p' | vedclass

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(B) Let the degree of the polynomial $p(x)$ be $n$. Then the degree of $p'(x)$ is $n-1$ and the degree of $p''(x)$ is $n-2$. Equating the degrees on b

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) Let the degree of the polynomial $p(x)$ be $n$. Then the degree of $p'(x)$ is $n-1$ and the degree of $p''(x)$ is $n-2$. Equating the degrees on both sides of the equation $p(2x) = p'(x) \cdot p''(x)$, we get: $n = (n-1) + (n-2) \Rightarrow n = 3$. Let $p(x) = ax^3$. Then $p'(x) = 3ax^2$ and $p''(x) = 6ax$. Substituting these into the given equation: $a(2x)^3 = (3ax^2) \cdot (6ax)$ $8ax^3 = 18a^2x^3$ Since $p(x)$ is a polynomial of degree $3$, $a 
eq 0$. $8a = 18a^2 \Rightarrow a = \frac{8}{18} = \frac{4}{9}$. Thus, $p(x) = \frac{4}{9}x^3$. Now, we calculate the sum: $\sum_{x=1}^5 p(x) = \frac{4}{9} \sum_{x=1}^5 x^3 = \frac{4}{9} \left( \frac{5(5+1)}{2} ight)^2$ $= \frac{4}{9} \left( \frac{30}{2} ight)^2 = \frac{4}{9} \cdot 15^2 = \frac{4}{9} \cdot 225 = 4 \cdot 25 = 100$.

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

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