AMC10 2019 10A, Problem 19

What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\]where $x$ is a real number?

Solution: Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $\fbox{2018}$.

Answer: 2018