Solve -8x^2+2x

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

Factor

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

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.

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

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-8\right)}$$

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.

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

Divide 0 by -16.

$$x=0$$

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

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

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

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

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 \frac{1}{4} for x_{2}.

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

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

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

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

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

Evaluate

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