Let f(n)=2n2+5n−3
By the rational roots theorem, if f(n)=0 has rational roots then they are all of the form pq in lowest terms, where p is a divisor of 3 and q is a divisor of 2.
Moreover, since 2 only factors as 1×2 (or −1×−2), one of the two corresponding linear factors must have q=±1, so pq is an integer.
As a result, one of ±1 or ±3 must be a root of f(n)=0...
f(1)=2+5−3=4
f(−1)=2−5−3=−6
f(3)=18+15−3=30
f(−3)=18−15−3=0
So n=−3 is a root and (n+3) is a factor.
The other factor must be (2n−1) in order that the coefficient of n2 is 2 and the constant term is −3 when these two factors are multiplied.
It's actually quicker to do than to write these words, but we find:
2n2+5n−3=(n+3)(2n−1)
