Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa+12$. To find $p$ and $q$, set up a system to be solved.
$$p+q=8$$ $$pq=1\times 12=12$$
Since $pq$ is positive, $p$ and $q$ have the same sign. Since $p+q$ is positive, $p$ and $q$ 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 $8$.
$$p=2$$ $$q=6$$
Rewrite $a^{2}+8a+12$ as $\left(a^{2}+2a\right)+\left(6a+12\right)$.
$$\left(a^{2}+2a\right)+\left(6a+12\right)$$
Factor out $a$ in the first and $6$ in the second group.
$$a\left(a+2\right)+6\left(a+2\right)$$
Factor out common term $a+2$ by using distributive property.
$$\left(a+2\right)\left(a+6\right)$$
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 $ax^{2}+bx+c=0$.
$$a^{2}+8a+12=0$$
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.
$$a=\frac{-8±\sqrt{8^{2}-4\times 12}}{2}$$
Square $8$.
$$a=\frac{-8±\sqrt{64-4\times 12}}{2}$$
Multiply $-4$ times $12$.
$$a=\frac{-8±\sqrt{64-48}}{2}$$
Add $64$ to $-48$.
$$a=\frac{-8±\sqrt{16}}{2}$$
Take the square root of $16$.
$$a=\frac{-8±4}{2}$$
Now solve the equation $a=\frac{-8±4}{2}$ when $±$ is plus. Add $-8$ to $4$.
$$a=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$a=-2$$
Now solve the equation $a=\frac{-8±4}{2}$ when $±$ is minus. Subtract $4$ from $-8$.
$$a=-\frac{12}{2}$$
Divide $-12$ by $2$.
$$a=-6$$
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 $x_{2}$.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$a^{2}+8a+12=\left(a+2\right)\left(a+6\right)$$
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 $u$.
$$r = -4 - u$$ $$s = -4 + u$$
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 $s$.
