Question

How do you solve `24x^3+18x^2-168x = 0`?

Answer 1

First off, notice that you can divide out `x`. That means one of the solutions is `x = 0`.

`0 = 24x^2 + 18x - 168`

Now you can divide by 3.

`0 = 8x^2 + 6x - 56`

You can't really factor it into integers though. The factors of 56 include 1, 2, 4, 7, 8, 14, 28, and 56. The ones that could work with `8x` and `x` to give `6` are `7` with `8` or `4` with `14`; the rest are too far apart. Regardless, neither of those give `6x` in the end. So:

`x = [-b pm sqrt(b^2 - 4ac)]/(2a)`

`= [-6 pm sqrt(36 - 4*8*-56)]/(16)`

`= [-6 pm sqrt(1828)]/(16)`

`= [-6 pm 2sqrt(457)]/(16)`

Overall:

`x = 0`
`x ~~ 2.3125`
`x ~~-3.0625`