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

1 Answer

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