Solve y^2+8y+12

$${ y }^{ 2 } +8y+12$$

Factor

$$\left(y+2\right)\left(y+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite y^{2}+8y+12 as \left(y^{2}+2y\right)+\left(6y+12\right).

$$\left(y^{2}+2y\right)+\left(6y+12\right)$$

Factor out y in the first and 6 in the second group.

$$y\left(y+2\right)+6\left(y+2\right)$$

Factor out common term y+2 by using distributive property.

$$\left(y+2\right)\left(y+6\right)$$

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.

$$y^{2}+8y+12=0$$

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.

$$y=\frac{-8±\sqrt{8^{2}-4\times 12}}{2}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 12}}{2}$$

Multiply -4 times 12.

$$y=\frac{-8±\sqrt{64-48}}{2}$$

Add 64 to -48.

$$y=\frac{-8±\sqrt{16}}{2}$$

Take the square root of 16.

$$y=\frac{-8±4}{2}$$

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

$$y=-\frac{4}{2}$$

Divide -4 by 2.

$$y=-2$$

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

$$y=-\frac{12}{2}$$

Divide -12 by 2.

$$y=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and -6 for x_{2}.

$$y^{2}+8y+12=\left(y-\left(-2\right)\right)\left(y-\left(-6\right)\right)$$

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

$$y^{2}+8y+12=\left(y+2\right)\left(y+6\right)$$

Evaluate

$$\left(y+2\right)\left(y+6\right)$$