Question

How do you factor `3x^3-x^2-3x-1` ?

Answer 1

With a change of sign this factors by grouping...

Explanation:

This question appears to have a wrong sign in it. As given it has one irrational zero and two complex zeros, but changing any one of the signs results in a cubic that factors by grouping:

`3x^3-x^2-3x-1` is given

`color(white)()`
Example one

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

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

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

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

`color(white)()`
Example two

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

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

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

`color(white)()`
Example three

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

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

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

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