Try this beautiful problem from Algebra based on Least Possible Value.
What is the least possible value of \(((x+1)(x+2)(x+3)(x+4)+2019)\)
where (x) is a real number?
But try the problem first...
AMC-10A (2019) Problem 19
Pre College Mathematics
To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)
Let us take \(((x^2+5x+5=m))\)
then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)
Can you now finish the problem ..........
Clearly in \((m^2+2018).......(m^2)\) is positive ( squares of any number is non-negative) and least value is 0
can you finish the problem........
Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .