Solve y^2+8y+12=

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

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.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +8x +12 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -8 rs = 12$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 12

$$(-4 - u) (-4 + u) = 12$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$16 - u^2 = 12$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 12-16 = -4$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-4 - 2 = -6 s = -4 + 2 = -2$$

Evaluate

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