Solve 9x^2-5x

$$9 x ^ { 2 } - 5 x$$

Factor

$$x\left(9x-5\right)$$

Solution Steps

Factor out x.

$$x\left(9x-5\right)$$

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.

$$9x^{2}-5x=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 9}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 9}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{5±5}{18}$$

Now solve the equation x=\frac{5±5}{18} when ± is plus. Add 5 to 5.

$$x=\frac{10}{18}$$

Reduce the fraction \frac{10}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{9}$$

Now solve the equation x=\frac{5±5}{18} when ± is minus. Subtract 5 from 5.

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

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{5}{9} for x_{1} and 0 for x_{2}.

$$9x^{2}-5x=9\left(x-\frac{5}{9}\right)x$$

Subtract \frac{5}{9} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-5x=9\times \frac{9x-5}{9}x$$

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

$$9x^{2}-5x=\left(9x-5\right)x$$

Evaluate

$$x\left(9x-5\right)$$