Solve (5x^2-5x)

$$( 5 x ^ { 2 } - 5 x )$$

Factor

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

Solution Steps

Factor out 5.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

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

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

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{5±5}{10}$$

Now solve the equation x=\frac{5±5}{10} when ± is plus. Add 5 to 5.

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

Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{5±5}{10} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

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

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

Evaluate

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