How do you factor #2x^2+ 11x + 5#?

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Answer 1 · 2015-05-28T19:42:01

I'll use a version of the AC Method.

The coefficients are #A=2#, #B=11#, #C=5# (ignoring any signs).

Since the sign on the constant term is #+#, look for a pair of factors of #AC=10#, whose sum is #11#. The pair #1, 10# works.

Then use that pair to split the middle term and factor by grouping:

#2x^2+11x+5 = 2x^2+x+10x+5#

#=(2x^2+x)+(10x+5)#

#=x(2x+1)+5(2x+1)#

#=(x+5)(2x+1)#