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Question

The four zeros of the polynomial x^4 + jx^2 + kx + 225 are distinct real numbers in arithmetic progression. Compute the value of j.

Accepted Answer

Let the four roots be a, a + d, a + 2d, and a + 3d. Then by Vieta's formulas, their sum is 0: 4a + 6d = 0.Then d = -23 a, so the four roots are a, a3, -a3, and -a. Their product is a a3 ( -a3 ) (-a) = a^49 = 225,so a = 3 5. Hence, the four roots are 3 5, 5, -5, -3 5, and the polynomial is (x - 3 5)(x - 5)(x + 5)(x + 3 5) = (x^2 - 5)(x^2 - 45) = x^4 - 50x^2 + 225.Thus, j = -50.