2022 AMC 10B Problem 7

If a quadratic $x^2 + kx + 36$ has roots $r$ and $s$, it factors as $(x - r)(x - s)$. Multiply out to compare:
$$(x - r)(x - s) = x^2 - (r + s)x + rs.$$
So $rs = 36$ and $k = -(r + s)$. The plan: list the pairs of integers that multiply to $36$, then read off $k$.

1. Positive pairs. $36 = 1 \times 36 = 2 \times 18 = 3 \times 12 = 4 \times 9 = 6 \times 6$.
2. Drop $6 \times 6$. The roots must be different, so the pair $6, 6$ is not allowed. Four pairs are left.
3. Their values of $k$. The sums are $37, 20, 15, 13$, so $k = -37, -20, -15, -13$.
4. Negative pairs. Two negative numbers also multiply to a positive number: $(-1)(-36)$, $(-2)(-18)$, $(-3)(-12)$, $(-4)(-9)$. Their sums are $-37, -20, -15, -13$, so $k = 37, 20, 15, 13$.
5. Mixed signs are impossible. One positive and one negative root would multiply to a negative number, not $36$.
6. Count. The eight values $\pm 13, \pm 15, \pm 20, \pm 37$ are all different.

The answer is $\boxed{\textbf{(B)}\ 8}$.

For $x^2 + kx + c$, the roots multiply to $c$ and add to $-k$ (these facts are called Vieta's formulas). So "integer roots" turns into "pairs of whole numbers that multiply to $c$", and each pair gives one value of $k$. Remember the negative pairs, throw out the pair with two equal numbers, and do not count a pair twice just because you can swap its order.