Solve 100x^2-20x

$$100 x ^ { 2 } - 20 x$$

Factor

$$20x\left(5x-1\right)$$

Solution Steps

Factor out 20.

$$20\left(5x^{2}-x\right)$$

Consider 5x^{2}-x. Factor out x.

$$x\left(5x-1\right)$$

Rewrite the complete factored expression.

$$20x\left(5x-1\right)$$

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.

$$100x^{2}-20x=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}}}{2\times 100}$$

Take the square root of \left(-20\right)^{2}.

$$x=\frac{-\left(-20\right)±20}{2\times 100}$$

The opposite of -20 is 20.

$$x=\frac{20±20}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{20±20}{200}$$

Now solve the equation x=\frac{20±20}{200} when ± is plus. Add 20 to 20.

$$x=\frac{40}{200}$$

Reduce the fraction \frac{40}{200} to lowest terms by extracting and canceling out 40.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{20±20}{200} when ± is minus. Subtract 20 from 20.

$$x=\frac{0}{200}$$

Divide 0 by 200.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{5} for x_{1} and 0 for x_{2}.

$$100x^{2}-20x=100\left(x-\frac{1}{5}\right)x$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$100x^{2}-20x=100\times \frac{5x-1}{5}x$$

Cancel out 5, the greatest common factor in 100 and 5.

$$100x^{2}-20x=20\left(5x-1\right)x$$

Evaluate

$$20x\left(5x-1\right)$$