i think you have to factor it first then find the solution(s). if you could explain this to me i would be so grateful!

## find all real number solutions: x^3-3x^2+2x=0

# 2 Answers

Although all cubic polynomials can be solved using radicals (Cardano's formula), in most middle and high school classes, all cubics that the teachers and textbooks ask you to solve have rational roots.

In this case, can you see that we can factor out x? this gives x(x^{2}-3x+2).

Also, recall how you factor quadratics using FOIL. In this case, the form is x^{2}-3x+2 = (x - __)(x - __).

We want the sum of the blanks to be 3 and the product to be 2. The only choice would be 1 and 2 (1+2=3 and 1*2 = 2).

So we have x^{3} - 3x^{2} + 2x = x(x-1)(x-2) = 0. I assume you can now figure out the roots.

the equation is

x^{3 }-3x^{2 }+2x= 0

taking x common on left hand side

x (x^{2 }-3x+2) = 0

x (x-2) (x-1) = 0

This is true for x= 0 or 2 or 1

Hence there are 3 real number solutions