2019 AMC 10A Problem 19

Group the factors so that each pair has the same $x$-coefficient after expanding:
$$(x+1)(x+4) = x^2 + 5x + 4,\qquad (x+2)(x+3) = x^2 + 5x + 6.$$
Let $u = x^2 + 5x$. The product becomes
$$(u + 4)(u + 6) = u^2 + 10u + 24 = (u + 5)^2 - 1 \ge -1.$$
Equality needs $u = -5$, i.e. $x^2 + 5x + 5 = 0$, which has real solutions since its discriminant $25 - 20 = 5$ is positive. So the product's least value is $-1$, and the least value of the whole expression is $-1 + 2019 = 2018$.

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

A product of four linear factors whose roots are symmetric about a center ($-2.5$ here) collapses to a quadratic in the symmetric variable $u = x^2 + 5x$; completing the square then reveals the minimum. Always check that the minimizing $u$ is actually attained by a real $x$, since otherwise the bound is not the true minimum.

Shift to the center: let $t = x + 2.5$. Then the product is $(t^2 - \tfrac94)(t^2 - \tfrac14)$, a quadratic in $t^2$ with minimum $-1$ at $t^2 = \tfrac54$. The symmetry makes the vertex obvious.

• Testing only integer values of x (which give a product of 0 or 24) and concluding the minimum is 2019.
• Grouping $(x+1)(x+2)$ with $(x+3)(x+4)$, which does not share a common quadratic part and leads nowhere.