Solve -4y^2-4y

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

Factor

$$-4y\left(y+1\right)$$

Solution Steps

Factor out 4.

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

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

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

Rewrite the complete factored expression.

$$4y\left(-y-1\right)$$

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.

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

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.

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$y=\frac{4±4}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{4±4}{-8}$$

Now solve the equation y=\frac{4±4}{-8} when ± is plus. Add 4 to 4.

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

Divide 8 by -8.

$$y=-1$$

Now solve the equation y=\frac{4±4}{-8} when ± is minus. Subtract 4 from 4.

$$y=\frac{0}{-8}$$

Divide 0 by -8.

$$y=0$$

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}.

$$-4y^{2}-4y=-4\left(y-\left(-1\right)\right)y$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-4y^{2}-4y=-4\left(y+1\right)y$$

Evaluate

$$-4y\left(y+1\right)$$