Solve (y^2+8y+4)

$$( { y }^{ 2 } +8y+4)$$

Factor

$$\left(y-\left(-2\sqrt{3}-4\right)\right)\left(y-\left(2\sqrt{3}-4\right)\right)$$

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.

$$y^{2}+8y+4=0$$

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.

$$y=\frac{-8±\sqrt{8^{2}-4\times 4}}{2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 4}}{2}$$

Multiply -4 times 4.

$$y=\frac{-8±\sqrt{64-16}}{2}$$

Add 64 to -16.

$$y=\frac{-8±\sqrt{48}}{2}$$

Take the square root of 48.

$$y=\frac{-8±4\sqrt{3}}{2}$$

Now solve the equation y=\frac{-8±4\sqrt{3}}{2} when ± is plus. Add -8 to 4\sqrt{3}.

$$y=\frac{4\sqrt{3}-8}{2}$$

Divide -8+4\sqrt{3} by 2.

$$y=2\sqrt{3}-4$$

Now solve the equation y=\frac{-8±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from -8.

$$y=\frac{-4\sqrt{3}-8}{2}$$

Divide -8-4\sqrt{3} by 2.

$$y=-2\sqrt{3}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+2\sqrt{3} for x_{1} and -4-2\sqrt{3} for x_{2}.

$$y^{2}+8y+4=\left(y-\left(2\sqrt{3}-4\right)\right)\left(y-\left(-2\sqrt{3}-4\right)\right)$$

Evaluate

$$y^{2}+8y+4$$