Solve (3x^2-8x)=

$$( 3 x ^ { 2 } - 8 x ) =$$

Factor

$$x\left(3x-8\right)$$

Solution Steps

Factor out x.

$$x\left(3x-8\right)$$

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.

$$3x^{2}-8x=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 3}$$

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

$$x=\frac{-\left(-8\right)±8}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±8}{6}$$

Now solve the equation x=\frac{8±8}{6} when ± is plus. Add 8 to 8.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{8±8}{6} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and 0 for x_{2}.

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

Subtract \frac{8}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

$$x\left(3x-8\right)$$