2019 AMC 10A Problem 24

Since $s^3 - 22s^2 + 80s - 67 = (s-p)(s-q)(s-r)$, multiply the identity by this product:
$$1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q).$$
This holds for all $s$ (both sides are polynomials that agree at infinitely many points). Substituting $s = p$ kills the $B$ and $C$ terms and leaves $1 = A(p-q)(p-r)$, so
$$\frac1A = (p-q)(p-r),\qquad \frac1B = (q-p)(q-r),\qquad \frac1C = (r-p)(r-q).$$

Add them. Expanding each product,
$$(p-q)(p-r) = p^2 - p(q+r) + qr,$$
and summing the three cyclic versions gives
$$p^2 + q^2 + r^2 - 2(pq + qr + rp) + (pq + qr + rp) = p^2 + q^2 + r^2 - (pq + qr + rp).$$
Using $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+rp)$, the total is
$$(p+q+r)^2 - 3(pq + qr + rp).$$
By Vieta, $p + q + r = 22$ and $pq + qr + rp = 80$, so the sum is $22^2 - 3 \cdot 80 = 484 - 240 = 244$.

The answer is $\boxed{\textbf{(B)}\ 244}$.

Partial-fraction constants are found by clearing denominators and plugging in a root (the "cover-up" method), which expresses each $1/A$ as a product of root differences. Any symmetric expression in the roots is then a combination of the Vieta sums; here it is $e_1^2 - 3e_2$, so the roots themselves are never needed and the constant term $67$ is irrelevant.

Recognize $(p-q)(p-r) = f'(p)$ where $f(x) = (x-p)(x-q)(x-r)$, so $\frac1A + \frac1B + \frac1C = f'(p) + f'(q) + f'(r)$. With $f'(x) = 3x^2 - 44x + 80$, this is $3(p^2+q^2+r^2) - 44 \cdot 22 + 240 = 3(484 - 160) - 968 + 240 = 244$.

• Trying to find the roots or the individual constants A, B, C numerically instead of expressing 1/A + 1/B + 1/C symmetrically and using Vieta.
• Sign slips when summing the cross terms (getting $(p+q+r)^2 - 2(pq+qr+rp) = 324$ or $-3 \cdot 80$ added instead of subtracted).