Question

What is the least possible value of (x +1)(x+2)(x+3)(x +4)+2019 where x is a real

number?
MANY POINTS

Answer 1

f(x)=(x+1)(x+2)(x+3)(x+4)+2019

f(x)=(x2+5x+4)(x2+5x+6)+2019

Suppose that y=x2+5x

Hence we have f(y)f(y)=(y+4)(y+6)+2019=y2+10y+24+2019=y2+10y+25+2018=(y+5)2+2018≥2018[∵(y+5)2≥0,∀y∈R]

and therefore…. min (f(x))=2018

ANSWER = 2018

Explanation:

hope that helps >3

Answer 2

2018

Explanation:

By grouping the first, last and two middle terms, we get (
`x^(2)`+5x+4)(
`x^(2)`+5x+6) + 2019. This can then be simplified to (
`x^(2)`+5x+2)^2 - 1 + 2019 Noting that squares are nonnegative, and verifying that
`x^(2)` + 5x + 5 = 0 for some real x, the answer is 2018.