Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

2. Polynomial
$p(x)=x^{3}-224 x^{2}+2016 x-d$

has three roots that form a geometric progression. Then the value of $d$ is $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solutions — 2

Solution 1

2.729.

Let the three roots of the polynomial be $a, b, c$ , and $b^{2}=a c$ .
By Vieta's formulas, we have
$\left\{\begin{array}{l} a+b+c=224, \\ a b+b c+c a=2016, \\ a b c=d . \end{array}\right.$

Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$ .
Thus, $d=a b c=b^{3}=729$ .

Solution 2

2. 729 .

Let the three roots of the polynomial be $a$ , $b$ , and $c$ , and $b^{2}=a c$ . By Vieta's formulas, we have
$\left\{\begin{array}{l} a+b+c=224, \\ a b+b c+c a=2016, \\ a b c=d . \end{array}\right.$

Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$ .
Thus, $d=a b c=b^{3}=729$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.