need help with factoring quadratic equations

x^2-3x+2=x(x-1)(x-2)
correction: x^2-3x+2 = (x-1)(x-2)
(you wrote an extra x)
anyway a quadratic can be written as (x-a)(x-b) (where a and b are roots/solution)
= x^2 - (a+b)x+ab
Thus in any quadratic -1*coefficient of x is sum of roots and the constant term is product of roots.
also discriminant, d =b^2 - 4ac
if d>0 ->distinct and real roots
d=0 means perfect square (i.e. same real roots)
d<0 complex roots
for equation ax^2 + bx + c,
roots are (-b+root(d)) /2a and (-b-root(d)) /2a )

ye you're right about an extra "x", my mistake.

The original equation was x^3-3x^2+2x so i factored x(x^2-3x+2)=x(x-1)(x-2) but I forgot to remove "x" from my post.

Thanks for your explanation.
I remember about the discriminant from school, but didn't remember about factoring to (x-...)(x-...)[/b]

the first thing you need to do is to break the middle term in 2 terms which equals to the multiplication of the first term & the Third term
lets see here
x^2-3x+2
x^2-2x-x+2
(x-2)(x-1) which is your answer

there are other methods also to solve these first is through Quadratic formula
which is x=(-b±?(b²-4ac))/(2a)
this will be helpful for imaginary solutions

the last method is to complete the square to get the solution

I hope you now know How to solve quadratic equations
If you still have some doubts check the link below

https://math.tutorvista.com/algebra/quad ... tions.html