2015 AMC 10A Problem 23

Turn "the zeros are whole numbers" into an equation about the zeros themselves.

1. Call the zeros $r$ and $s$. Then $x^2 - ax + 2a = (x - r)(x - s) = x^2 - (r + s)x + rs$. Matching the terms:
$$r + s = a, \qquad rs = 2a.$$
2. Get rid of $a$: the product is twice the sum,
$$rs = 2r + 2s.$$
3. This almost factors. Since $(r - 2)(s - 2) = rs - 2r - 2s + 4$, move everything to one side and add $4$:
$$(r - 2)(s - 2) = 4.$$
4. Now $r - 2$ and $s - 2$ are whole numbers that multiply to $4$. List every pair, including negative ones:

(Swapping the order, like $4,\ 1$, gives the same zeros and the same $a$.)
5. Check each one works:
- $a = 9$: $x^2 - 9x + 18 = (x - 3)(x - 6)$.
- $a = 8$: $x^2 - 8x + 16 = (x - 4)^2$.
- $a = -1$: $x^2 + x - 2 = (x + 2)(x - 1)$.
- $a = 0$: the function is $x^2$, whose zero $0$ is a whole number.
6. The sum of the possible values is $9 + 8 + (-1) + 0 = 16$.

The answer is $\boxed{\textbf{(C)}\ 16}$.

"The zeros are integers" together with sum-and-product of the zeros turns a quadratic into a whole-number puzzle. An equation like $rs = 2(r + s)$ becomes a product by adding a constant, $(r - 2)(s - 2) = 4$, and then only a short list of factor pairs is possible. Always include negative pairs and equal pairs (a repeated zero).

A quadratic with integer coefficients has whole-number zeros only if the number under the square root in the quadratic formula, $a^2 - 8a$, is a perfect square $k^2$. Completing the square, $(a - 4)^2 - k^2 = 16$, so
$$(a - 4 - k)(a - 4 + k) = 16.$$
The two factors differ by an even number, so both are even: $2 \times 8$, $4 \times 4$, $(-8) \times (-2)$ or $(-4) \times (-4)$. Their average, $a - 4$, is $5, 4, -5, -4$, so $a = 9, 8, -1, 0$, with sum $16$.