How do you factor #5v ^ { 2} - 15v - 90# completely?

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Answer 1 · 2018-04-20T06:42:36

#5(v-6)(v+3)#

#5v^2 - 15v - 90=5(v^2 - 3v - 18)#

We want to find two numbers that when added give us #3# and when multiplied give us #-18#.

The factors of #18# are #2,3,6,# and #9#.

#6-3=3#
#6 times -3=-18#

The numbers #6# and #-3# satisfy these two equations.

#v^2 - 3v - 18 = (v-6)(v+3)#

We can double check this by expanding again.
#(v-6)(v+3) = v(v+3)-6(v+3)= v^2 + 3v -6v -18 = v^2 -3v -18#

Multiplying all of this by five we get

#5(v-6)(v+3)#

Answer 2 · 2018-04-20T07:41:04

#5(v-6)(v+3)#

#"take out a "color(blue)"common factor "5#

#=5(v^2-3v-18)#

#"the factors of - 18 which sum to - 3 are - 6 and + 3"#

#=5(v-6)(v+3)#