Question

What is the least possible value of $\begin{aligned} &(x+1)(x+2)(x+3)(x+4)\\ &\quad+2019, \end{aligned}$ where $x$ is a real number?

$2017$

$2018$

$2019$

$2020$

$2021$

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.