Polynomials

Let’s say the polynomial has roots $p$ , $q$ , and $r$ . By Vieta’s Formulas, the sum of the roots $p + q + r = \frac{-(-2a)}{2} = a$ . Similarly, the sum of the products of each pair of roots $pq + pr + qr = \frac{a^2 - 81}{2}$ . Since $a = p + q + r$ , $a^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2qr$ . We can eliminate the $2pq + 2pr + 2qr$ since it equals $a^2 - 81$ : $a^2 = p^2 + q^2 + r^2 + a^2 - 81$ $\rightarrow$ $p^2 + q^2 + r^2 = 81$ . Without Loss of Generality, suppose that $p \ge q \ge r$ . Trial and error gives the triples $(p, q, r)$ : $(8, 4, 1), (7, 4, 4), (6, 6, 3)$ . Since there is a given value of $a$ , the two triples selected must be $(7, 4, 4)$ and $(6, 6, 3)$ because $7 + 4 + 4 = 15 = 6 + 6 + 3$ . $a$ directly follows to be $15$ . But we need to find the sum of the values of $c$ : Vieta’s formulas again show that $c = 2\cdot \prod \text{(roots)}$ . The requested sum is therefore $2 \cdot [7 \cdot 4 \cdot 4 + 6 \cdot 6 \cdot 3] = 2 \cdot (112 + 108) = 2 \cdot 220 = \boxed{440}$ .

I thought this is problem 6 of 2015 AIME II.