Solve 4y-2y^2

$$4y-2 { y }^{ 2 }$$

Factor

$$2y\left(2-y\right)$$

Solution Steps

Factor out 2.

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

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

$$y\left(2-y\right)$$

Rewrite the complete factored expression.

$$2y\left(-y+2\right)$$

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.

$$-2y^{2}+4y=0$$

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.

$$y=\frac{-4±\sqrt{4^{2}}}{2\left(-2\right)}$$

Take the square root of 4^{2}.

$$y=\frac{-4±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{-4±4}{-4}$$

Now solve the equation y=\frac{-4±4}{-4} when ± is plus. Add -4 to 4.

$$y=\frac{0}{-4}$$

Divide 0 by -4.

$$y=0$$

Now solve the equation y=\frac{-4±4}{-4} when ± is minus. Subtract 4 from -4.

$$y=-\frac{8}{-4}$$

Divide -8 by -4.

$$y=2$$

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 2 for x_{2}.

$$-2y^{2}+4y=-2y\left(y-2\right)$$

Evaluate

$$2y\left(2-y\right)$$