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

Hint: In this question we will simplify the polynomial by taking out the common term $x$, then we will solve the simplified quadratic equation and find the factors of the equation, which are the multiplication of linear terms, which will give us the required values of $x$.

Formula used: $({x_1},{x_2}) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where $({x_1},{x_2})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.

Complete step-by-step solution:
We have the given equation as:
 $({x^3} + {x^2} - x)$
Now since the term $x$ is common in all the terms, we will take it out as common, on taking it out as common, we get:
$\Rightarrow$$x({x^2} + x - 1)$
Now, we use the term $({x^2} + x - 1)$ in the form of a quadratic equation, therefore we solve it by splitting the middle term. Now in this equation, splitting the middle term is not possible therefore, we will use the quadratic formula.
In this question we have:
$a = 1$
$b = 1$
$c = - 1$
On substituting the values in the formula, we get:
$\Rightarrow$$({x_1},{x_2}) = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 1)} }}{{2(1)}}$
On simplifying the root, we get:
$\Rightarrow$\[({x_1},{x_2}) = \dfrac{{ - 1 \pm \sqrt {1 + 4} }}{2}\]
On simplifying, we get:
$\Rightarrow$\[{x_1} = \dfrac{{ - 1 + \sqrt 5 }}{2}\] and \[{x_2} = \dfrac{{ - 1 - \sqrt 5 }}{2}\]

Therefore, the required roots of the equation are: $0$ , \[\dfrac{{ - 1 + \sqrt 5 }}{2}\] and \[\dfrac{{ - 1 - \sqrt 5 }}{2}\].

Note: It is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $3rd$ degree polynomial equation.
The factors of a polynomial equation are the terms of degree $1$ , which have to be multiplied among themselves so that we get the required polynomial equation. In the question above, $x$ , and are factors of the given polynomial.
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore a quadratic formula is used to solve these types of questions, the quadratic formula is used.