If f(x) is a polynomial such that f(x) = [f'(x)]^2 and f(2) = 0 , then f(-2) =

Step-by-step solution

Let the degree of the polynomial f(x) be n . The degree of f'(x) is n-1 , so the degree of [f'(x)]^2 is 2(n-1) . Equating the degrees, n = 2n - 2 n = 2 . Let f(x) = ax^2 + bx + c , where a 0 . Then f'(x) = 2ax + b . Substituting into the given equation f(x) = [f'(x)]^2 : ax^2 + bx + c = (2ax + b)^2 ax^2 + bx + c = 4a^2x^2 + 4abx + b^2 Comparing the coefficients of like powers of x : a = 4a^2 a = 1 4 (since a 0 ) b = 4ab b = b c = b^2 Thus, the polynomial is f(x) = 1 4 x^2 + bx + b^2 . Given f(2) = 0 : 1 4 (2)^2 + 2