Solve 8x^2-2x-0

$$8 { x }^{ 2 } -2x-0$$

Evaluate

$$2x\left(4x-1\right)$$

Solution Steps

Multiply -1 and 0 to get 0.

$$8x^{2}-2x+0$$

Anything plus zero gives itself.

$$8x^{2}-2x$$

Factor

$$2x\left(4x-1\right)$$

Solution Steps

Factor out 2.

$$2\left(4x^{2}-x\right)$$

Consider 4x^{2}-x. Factor out x.

$$x\left(4x-1\right)$$

Rewrite the complete factored expression.

$$2x\left(4x-1\right)$$

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.

$$8x^{2}-2x=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}}}{2\times 8}$$

Take the square root of \left(-2\right)^{2}.

$$x=\frac{-\left(-2\right)±2}{2\times 8}$$

The opposite of -2 is 2.

$$x=\frac{2±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{2±2}{16}$$

Now solve the equation x=\frac{2±2}{16} when ± is plus. Add 2 to 2.

$$x=\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{2±2}{16} when ± is minus. Subtract 2 from 2.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and 0 for x_{2}.

$$8x^{2}-2x=8\left(x-\frac{1}{4}\right)x$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-2x=8\times \frac{4x-1}{4}x$$

Cancel out 4, the greatest common factor in 8 and 4.

$$8x^{2}-2x=2\left(4x-1\right)x$$