How do you factor #2v^2-4-7v #?

1 Answer
Mar 1, 2017

Answer:

#2v^2-4-7v = (2v+1)(v-4)#

Explanation:

First put in standard order of decreasing powers of #v#:

#2v^2-7v-4#

Next, use an AC method:

Find a pair of factors of #AC=2*4=8# which differ by #B=7#.

The pair #8, 1# works.

Use this pair to split the middle term and factor by grouping:

#2v^2-7v-4 = 2v^2-8v+v-4#

#color(white)(2v^2-7v-4) = (2v^2-8v)+(v-4)#

#color(white)(2v^2-7v-4) = 2v(v-4)+1(v-4)#

#color(white)(2v^2-7v-4) = (2v+1)(v-4)#