Solve 6x^2-8x

$$6 { x }^{ 2 } -8x$$

Factor

$$2x\left(3x-4\right)$$

Solution Steps

Factor out 2.

$$2\left(3x^{2}-4x\right)$$

Consider 3x^{2}-4x. Factor out x.

$$x\left(3x-4\right)$$

Rewrite the complete factored expression.

$$2x\left(3x-4\right)$$

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.

$$6x^{2}-8x=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 6}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 6}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{8±8}{12}$$

Now solve the equation x=\frac{8±8}{12} when ± is plus. Add 8 to 8.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{8±8}{12} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and 0 for x_{2}.

$$6x^{2}-8x=6\left(x-\frac{4}{3}\right)x$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-8x=6\times \frac{3x-4}{3}x$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6x^{2}-8x=2\left(3x-4\right)x$$

Evaluate

$$2x\left(3x-4\right)$$