Question

How do you factor `6x^3-9x+3`?

Answer 1

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

Explanation:

`6*x^3-9 x+3=3(2*x^2-3*x+1)` but `2*x^2-3*x+1` has `x-1` as factor because `2*(1)^2-3*(1)+1 = 0` so
`(2*x^3-3*x+1)=(x-1)*(a*x^2+b*x+c)`
doing the product and equating coefficients
`2*x^3+0*x^2-3*x+1 = a*x^3+(b-a)*x^2+(c-b)*x-c`
then
`((2=a),(0=(b-a)),(-3=(c-b)),(1=-c))`
solving `a = 2, b = 2,c = -1` so
`(2*x^3-3*x+1)=(x-1)*(2*x^2+2*x-1)`
but `(2*x^2+2*x-1)=2(x+(1 +sqrt(3))/2)(x+(1 -sqrt(3))/2)`
Finally
`6*x^3-9 x+3=3*2*(x-1)(x+(1 +sqrt(3))/2)(x+(1 -sqrt(3))/2)`