Question

Is it possible to factor `y= 6x^3-9x+3`? If so, what are the factors?

Answer 1

Yes:

`y = 6x^3-9x+3`

`=3(x-1)(2x^2+2x-1)`

`= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this later.

First separate out the common scalar factor `3` to find:

`y = 6x^3-9x+3 = 3(2x^3-3x+1)`

Next note that the sum of the coefficients is zero. That is `2-3+1 = 0`. So `x=1` is a zero and `(x-1)` is a factor:

`3(2x^3-3x+1) = 3(x-1)(2x^2+2x-1)`

We can factor the remaining quadratic expression by completing the square and using the difference of squares identity...

`(2x^2-2x-1)`

`=2(x^2-x-1/2)`

`=2(x^2-x+1/4-3/4)`

`=2((x-1/2)^2-(sqrt(3)/2)^2)`

`=2((x-1/2)-sqrt(3)/2)((x-1/2)+sqrt(3)/2)`

`=2(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)`

Putting it all together:

`y = 6x^3-9x+3`

`=3(x-1)(2x^2+2x-1)`

`= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)`