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Question

Given that three roots of f(x)=x^4+ax^2+bx+c are 2, -3, and 5, what is the value of a+b+c?

Accepted Answer

By Vieta's formulas, the sum of the roots is 0, so the fourth root must be -4. Hence, f(x) = (x - 2)(x + 3)(x - 5)(x + 4).Then f(1) = (1 - 2)(1 + 3)(1 - 5)(1 + 4) = 80. But f(1) = 1 + a + b + c, so a + b + c = 79.