jmc

Question

The roots of x^3 + Px^2 + Qx - 19 = 0 are each one more than the roots of x^3 - Ax^2 + Bx - C = 0. If A, B, C, P, and Q are constants, compute A + B + C.

Accepted Answer

Let r, s, t be the roots of x^3 + Px^2 + Qx - 19 = 0. Let u = r - 1, v = s - 1, and w = t - 1, so u, v, w are the roots of x^3 - Ax^2 + Bx - C = 0. Thus, x^3 - Ax^2 + Bx - C = (x - u)(x - v)(x - w).Settting x = -1, we get -1 - A - B - C = (-1 - u)(-1 - v)(-1 - w) = -(u + 1)(v + 1)(w + 1) = -rst.By Vieta's formulas, rst = 19, so -rst = -19. Hence, -1 - A - B - C = -19.Then A + B + C = 19 - 1 = 18.