Question

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

Answer 1

`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))`

Answer 2

`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)`

Answer 3

`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)`