Question

How do you factor `3x^2-3x-8`?

Answer 1

`3x^2-3x-8` cannot be factored using integers.

You can try:
using `(3x+a)(x+b)` where `ab=-8` and `a+3b= -3`

Or using the product `3*-8 = -24` now look for two factors of `-24` that add to give you `-3`
(try `1* - 24`, `2* - 12`, `3* - 8`, `4* - 6` but none of them work.)

Using non-integers
Using the factor theorem allows us to factor the expression after we find its zeros:

Solve `3x^2-3x-8 = 0` by using the quadratic formula to get:

`x= (3+-sqrt 105)/6`

We can factor:
`3x^2-3x-8 = (x-(3+sqrt 105)/6)(x-(3-sqrt 105)/6)`.