Factor each polynomial by removing the common monomial factor. x^{3}-x^{2}+x

Step 1
The monomial factor is a factor which consist only single term . We call an expression binomial if it consist 2 terms and trinomial if it consists three terms . Here we get a trinomial expression by removing common monomial factor .
Then we have to find the factors of the resulting trinomial. we know that the general form of a trinomial is ${\displaystyle a{x}^2+bx+c=0}$
Step 2
We have a polynomial ${\displaystyle x^3-x^2+x}$, here we have to remove the common monomial factor .
We can clearly identify that all the terms in the given polynomial ${\displaystyle x^3-x^2+x}$ is a multiple of x , therefor we can take x as a common monomial factor
Thus we have ${\displaystyle x^3-x^2+x=x(x^2-x+1)}$
Now we have to find the factors of ${\displaystyle x^2-x+1}$ the trinomial.
By trinomial factorization method , to factorize ${\displaystyle x^2-x+1}$ we have to look for two integers a and b such that sum of a and b equal to the the coefficient of x (-1) and the product of a and b equal to the constant number (1) .
We know that only two integers gives the product value 1 is 1 and 1 itself. and the same time we cannot have -1 as a sum of two this kind of numbers.
Now we can use Discriminant to check the existence real roots or factors
If the discriminant ${\displaystyle b^2-4ac>0}$, then we have real roots
If the discriminant ${\displaystyle b^2-4ac<0}$, then we do not have real roots ( we may get imaginary roots )
We have trinomial ${\displaystyle x^2-x+1}$, where a=1, b=-1, c=1
${\displaystyle b^2-4ac={(-1)}^2-4\left(1\right)}$
=1-4
${\displaystyle b^2-4ac=-3}$
Thus we have discriminant ${\displaystyle b^2-4ac<0}$, hence we do not have any real roots or factors for ${\displaystyle x^2-x+1}$.
Hence the factor of the polynomial ${\displaystyle x^3-x^2+xisx(x^2-x+1)}$.