Solve f(x)=x-x^2

$$f ( x ) = x - x ^ { 2 }$$

Factor

$$x\left(1-x\right)$$

Solution Steps

Factor out x.

$$x\left(1-x\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-1\right)}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±1}{-2}$$

Now solve the equation x=\frac{-1±1}{-2} when ± is plus. Add -1 to 1.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-2} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 1 for x_{2}.

$$-x^{2}+x=-x\left(x-1\right)$$

Evaluate

$$x\left(1-x\right)$$