How do you factor #3v ^ { 2} + 12v - 63#?

3 Answers
Alan P.
Mar 20, 2018

#3v^2+12v-63=color(blue)(3(v+7)(v-3))#

Explanation:

Given
#color(white)("XXX")3v^2+12v-63#

First factor out the obvious constant
#color(white)("XXX")=color(red)3(v^2+4v-21)#

Then look for two numbers whose product is #(-21)# and whose sum is #(+4)#.
With a bit of playing around you should come up with #color(lime)(+7)# and #color(green)(-4)#;
enabling the continuation of the factoring:
#color(white)("XXX")color(red)3(vcolor(lime)(+7))(vcolor(green)(-4))#

George C.
Mar 20, 2018

#3x^2+12v-63 = 3(v-7)(v+3)#

Explanation:

Given:

#3x^2+12v-63#

First note that all of the coefficients are divisible by #3#, so we can separate that out as a factor:

#3v^2+12v-64 = 3(v^2-4v-21)#

To factor the remaining quadratic, we can find a pair of factors of #21# which differ by #4#. Note that #7 * 3 = 21# and #7 - 3 = 4#.

Hence we find:

#v^2-4v-21 = (v-7)(v+3)#

So:

#3x^2+12v-63 = 3(v-7)(v+3)#

Rosa.parks.history · NJ
Mar 20, 2018

#3(v+7)(v-4) #

Explanation:

First you take #3# out of all the number:
#3(v^2+ 4v - 21)#

Then factorise the middle:
so #7-3= 4#
and #7 xx - 3=-21#

so the final answer would be

#3(v+7)(v-4) #

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