2019 AMC 10A Problem 19

Multiplying the first two terms and the last terms yields $$(x^2 + 5x + 4)(x^2 + 5x + 6).$$ Note that these two terms differ by $2.$ We can try to express this as a difference of squares, which is $$(x^2 + 5x + 5)^2 - 1.$$ Adding $2019$ to this gets us $$(x^2 + 5x + 5)^2 + 2018.$$ Squares are non-negative, so as long as we find a way to make the inner expression $0,$ we can make the square $0.$ The discriminant is $5^2 - 4 \cdot 5 = 5,$ which is positive meaning that there is a value that makes the square $0.$ This means that the minimum value would be $$0^2 + 2018 = 2018.$$ Thus, B is the correct answer.