2019 AMC 10A Problem 19

Step 1: try a few values to see what is going on. At $x = -1, -2, -3$ or $-4$ one factor is $0$, so the product is $0$. Between those values it can dip below $0$. For example, at $x = -1.5$ the factors are $-0.5$, $0.5$, $1.5$, $2.5$, and the product is $-0.9375$. Guessing will not find the lowest point exactly, so we need a better way.

Step 2: pair the factors cleverly. Multiply the outer two and the inner two:
$$(x+1)(x+4) = x^2 + 5x + 4, \qquad (x+2)(x+3) = x^2 + 5x + 6.$$
Both start with $x^2 + 5x$. That is why this pairing is chosen.

Step 3: give the shared part a name. Let $u = x^2 + 5x$. The product becomes
$$(u + 4)(u + 6) = u^2 + 10u + 24.$$

Step 4: complete the square. $u^2 + 10u + 25 = (u + 5)^2$, so
$$u^2 + 10u + 24 = (u + 5)^2 - 1.$$
A square is never negative, so this is at least $-1$, and it equals $-1$ exactly when $u = -5$.

Step 5: check that $u = -5$ really happens. We need $x^2 + 5x = -5$, that is, $x^2 + 5x + 5 = 0$. Its discriminant is $5^2 - 4 \cdot 5 = 5$, which is positive, so there is a real $x$ that works.

Step 6: add 2019. The smallest value of the product is $-1$, so the smallest value of the whole expression is $-1 + 2019 = 2018$.

The answer is $\boxed{\textbf{(B)}\ 2018}$.

The four roots $-1, -2, -3, -4$ are evenly spaced, so pairing the outermost two and the innermost two gives two quadratics that differ only in their last number. Naming the shared part turns a degree-4 problem into a plain quadratic, and completing the square shows the lowest value. Always check that the lowest point can actually be reached with a real $x$.

Move the middle of the four roots to $0$. The roots are balanced around $-2.5$, so let $t = x + 2.5$. Then the factors are $t - 1.5$, $t - 0.5$, $t + 0.5$, $t + 1.5$, and the product is
$$(t^2 - 2.25)(t^2 - 0.25).$$
With $s = t^2$, this is $s^2 - 2.5s + 0.5625$, a quadratic whose lowest point is at $s = 1.25$, where it equals $1.5625 - 3.125 + 0.5625 = -1$. Again the answer is $2019 - 1 = 2018$.