Question

How do you solve `3x^3 -2x^2 -12x + 8 = 0`?

Answer 1

With a large polynomial like this, you'll need to factor in order to isolate the x and solve for it.

Explanation:

This is the equation you're given: `3x^3-2x^2-12x+8=0`

Separate the polynomial into two parts, and factor that way.
`(3x^3-2x^2)+(-12x+8)=0`
`x^2(3x-2)-4(3x-2)`
`(3x-2)(x^2-4)=0`

You can factor out `(x^2-4)` to `(x+2)(x-2)`.

The fully factored polynomial is `(3x-2)(x+2)(x-2)`.

But remember... the question asks you to solve the equation.
Set the factored polynomial to zero and solve for x (you will end up with three values of x):
`(3x-2)(x+2)(x-2)=0`

`3x-2=0` ---> `x=2/3`
`x+2=0` ---> `x=-2`
`x-2=0` ---> `x=2`

Your final answers are `x=2/3,-2,2`.

Answer 2

`x=-2`

`x=2`

`x=2/3`

Explanation:

`3x^3 -2x^2 -12x +8 =0`

Common factors

`3x(x+2)(x-2)-2(x+2)(x-2)=0`

`(3x-2)(x+2)(x-2) = 0`

Three solutions (B.S`=`both sides)

(1) `" "x+2=0" "` (`-2` B.S)

`x=-2`

(2) `" "x-2=0" "` (`+2` B.S)

`x=2`

(3) `" "3x-2=0" "` (`+2` B.S)

`3x=2" "` (`div3` B.S)

`x=2/3`

So, the problem has 3 solutions, `x=-2`, `x=2`, and `x=2/3`.

-Fish