Solve 5x-5x^2=

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

Factor

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

Solution Steps

Factor out 5.

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

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

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

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.

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

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-5\right)}$$

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.

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

Divide 0 by -10.

$$x=0$$

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

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

Divide -10 by -10.

$$x=1$$

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

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

Evaluate

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