Roots of the equation $f(x)=x^6-12x^5+bx^4+cx^3+dx^2+ex+64=0$ are positive. Which of the following has the greatest absolute value?
- $b$
- $c$
- $d$
- $e$
By Vieta's formulas for x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64 = 0, the product of the roots is 64. Since there are 6 positive roots, their geometric mean is (64)^(1/6) = 2. If all roots are 2, the polynomial is (x-2)^6 = x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 192x + 64. Comparing coefficients: b=60, c=-160, d=240, e=-192. The absolute values are 60, 160, 240, 192. The greatest is 240 (d).
Let the positive roots of f(x) be a, b, c, d, e, and g. By Vieta's formulas, their product abcdeg equals 64 and their sum abcdef equals 12. The coefficient b is -the sum of products of two roots, c is the sum of products of three roots, d is -the sum of products of four roots, and e is the sum of products of five roots. By Maclaurin's inequality for positive real numbers, the sum of products of k roots divided by the combination of 6 choose k is strictly decreasing. Therefore, the sum of products of 4 roots is strictly greater than the sum of products of 5 roots, meaning the absolute value of d is greater than the absolute value of e. Similarly comparing all coefficients shows that d has the greatest absolute value.