let's factor it ...
2n^3+n^2+6n+3
= n^2(2n+1) + 3(2n+1)
= (2n+1)(n^2 + 3)
what kind of a number is 2n+1 ??
Show that a function f(n)= 2n^3+n^2+6n+3 always produces a number that is divisible by an odd number greater than 1, for any natural number, n.
3 answers
I do not know.
Well, if n is a whole number, then 2n will always be even.
When you add 1 to any even, the result will be odd.
So 2n+1 is always odd
by writing it as
2n^3+n^2+6n+3
= (2n+1)(n^2 + 3)
I have shown that one of the factors, namely (2n+1), will always be an odd number, thus
2n^3+n^2+6n+3
is always divisible by an odd number
When you add 1 to any even, the result will be odd.
So 2n+1 is always odd
by writing it as
2n^3+n^2+6n+3
= (2n+1)(n^2 + 3)
I have shown that one of the factors, namely (2n+1), will always be an odd number, thus
2n^3+n^2+6n+3
is always divisible by an odd number
1 answer