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Question

Let a, b, c be the roots of the cubic x^3 + 3x^2 + 5x + 7 = 0. Given that P(x) is a cubic polynomial such that P(a) = b + c, P(b) = a + c, P(c) = a + b, and P(a + b + c) = -16, find P(x).

Accepted Answer

By Vieta's formulas, a + b + c = -3, so P(-3) = -16. Let Q(x) = P(x) + x + 3. Then align* Q(a) &= b + c + a + 3 = 0, Q(b) &= a + c + b + 3 = 0, Q(c) &= a + b + c + 3 = 0, Q(-3) &= P(-3) - 3 + 3 = -16. align*Hence, Q(x) = k(x - a)(x - b)(x - c) = k(x^3 + 3x^2 + 5x + 7) for some constant k. Setting x = -3, we get -16 = -8k,so k = 2. Then Q(x) = 2(x^3 + 3x^2 + 5x + 7), so P(x) = Q(x) - x - 3 = 2(x^3 + 3x^2 + 5x + 7) - x - 3 = 2x^3 + 6x^2 + 9x + 11.