Solve f(x)=2x^2-x

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

Factor

$$x\left(2x-1\right)$$

Solution Steps

Factor out x.

$$x\left(2x-1\right)$$

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.

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

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.

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±1}{4}$$

Now solve the equation x=\frac{1±1}{4} when ± is plus. Add 1 to 1.

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{1±1}{4} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

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

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

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-x=2\times \frac{2x-1}{2}x$$

Cancel out 2, the greatest common factor in 2 and 2.

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

Evaluate

$$x\left(2x-1\right)$$