Let p(x) be a real polynomial of least degree which has a local maximum at x=1 and a local minimum at x=3. If p(1)=6 and p(3)=2, then p'(0) is

Given that P(x) is a real polynomial of least degree. It has two extreme: local maximum at x=1 and local minimum at x=3. Hence, P'(x) = k(x-1)(x-3) = k(x² - 4x + 3) ⇒ p(x) = k(x³/3 - 2x² + 3x) + c
Given that P(1) = 6 ⇒ 4/3k + c = 6
Also, P(3) = 2 ⇒ c = 2
So, k = 3
∴ P'(0) = 3k = 9.