How do you factor #2n^2+5n-3#?

1 Answer
Jun 15, 2015

Answer:

#2n^2+5n-3 = (n+3)(2n-1)#

Explanation:

Let #f(n) = 2n^2+5n-3#

By the rational roots theorem, if #f(n) = 0# has rational roots then they are all of the form #p/q# in lowest terms, where #p# is a divisor of #3# and #q# is a divisor of #2#.

Moreover, since #2# only factors as #1 xx 2# (or #-1 xx -2#), one of the two corresponding linear factors must have #q = +-1#, so #p/q# 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 #n^2# 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:

#2n^2+5n-3 = (n+3)(2n-1)#