Solve 25x-5x^2

$$25 x - 5 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out 5.

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

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

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

Rewrite the complete factored expression.

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

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.

$$-5x^{2}+25x=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}}}{2\left(-5\right)}$$

Take the square root of 25^{2}.

$$x=\frac{-25±25}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-25±25}{-10}$$

Now solve the equation x=\frac{-25±25}{-10} when ± is plus. Add -25 to 25.

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

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-25±25}{-10} when ± is minus. Subtract 25 from -25.

$$x=-\frac{50}{-10}$$

Divide -50 by -10.

$$x=5$$

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 5 for x_{2}.

$$-5x^{2}+25x=-5x\left(x-5\right)$$

Evaluate

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