Solve 9x^2-4x

$$9 x ^ { 2 } - 4 x$$

Factor

$$x\left(9x-4\right)$$

Solution Steps

Factor out x.

$$x\left(9x-4\right)$$

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.

$$9x^{2}-4x=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 9}$$

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\times 9}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{4±4}{18}$$

Now solve the equation x=\frac{4±4}{18} when ± is plus. Add 4 to 4.

$$x=\frac{8}{18}$$

Reduce the fraction \frac{8}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{9}$$

Now solve the equation x=\frac{4±4}{18} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{9} for x_{1} and 0 for x_{2}.

$$9x^{2}-4x=9\left(x-\frac{4}{9}\right)x$$

Subtract \frac{4}{9} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-4x=9\times \frac{9x-4}{9}x$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}-4x=\left(9x-4\right)x$$

Evaluate

$$x\left(9x-4\right)$$