Question

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

Answer 1

The sum of all coefficients of given cubic equation: `x^3-3x^2-x+3=0` is zero hence `x=1` is a root of given cubic equation i.e. `(x-1)` is a factor of `x^3-3x^3-x+3`
Now, cubic algebraic polynomial: `x^3-3x^2-x+3` can be factorized as follows
`x^3-3x^3-x+3`
`=x^2(x-1)-2x(x-1)-3(x-1)`
`=(x-1)(x^2-2x-3)`
`=(x-1)(x^2-3x+x-3)`
`=(x-1)(x(x-3)+(x-3))`
`=(x-1)(x-3)(x+1)`
hence, the solution of given cubic equation will be given as
`(x-1)(x-3)(x+1)=0`
`x=-1, 1, 3`