Solve 5y^2-405y

$$5 y ^ { 2 } - 405 y$$

Factor

$$5y\left(y-81\right)$$

Solution Steps

Factor out 5.

$$5\left(y^{2}-81y\right)$$

Consider y^{2}-81y. Factor out y.

$$y\left(y-81\right)$$

Rewrite the complete factored expression.

$$5y\left(y-81\right)$$

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.

$$5y^{2}-405y=0$$

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.

$$y=\frac{-\left(-405\right)±\sqrt{\left(-405\right)^{2}}}{2\times 5}$$

Take the square root of \left(-405\right)^{2}.

$$y=\frac{-\left(-405\right)±405}{2\times 5}$$

The opposite of -405 is 405.

$$y=\frac{405±405}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{405±405}{10}$$

Now solve the equation y=\frac{405±405}{10} when ± is plus. Add 405 to 405.

$$y=\frac{810}{10}$$

Divide 810 by 10.

$$y=81$$

Now solve the equation y=\frac{405±405}{10} when ± is minus. Subtract 405 from 405.

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

Divide 0 by 10.

$$y=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 81 for x_{1} and 0 for x_{2}.

$$5y^{2}-405y=5\left(y-81\right)y$$

Evaluate

$$5y\left(y-81\right)$$