How do you factor #2v ^ { 2} - 9v - 11#?

1 Answer
George C.
Aug 8, 2017

#2v^2-9v-11 = (v+1)(2v-11)#

Explanation:

Given:

#2v^2-9v-11#

If you looked at this and thought "AC method", then you would be looking for a pair of factors of #AC=2*11 = 22# which differ by #B=9#.

If so, then they're right under your nose, namely #11# and #2#.

We can use this pair to split the middle term and factor by grouping:

#2v^2-9v-11 = (2v^2-11v)+(2v-11)#

#color(white)(2v^2-9v-11) = v(2v-11)+1(2v-11)#

#color(white)(2v^2-9v-11) = (v+1)(2v-11)#

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