Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa-6$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-1$$ $$pq=1\left(-6\right)=-6$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-6$.
$$1,-6$$ $$2,-3$$
Calculate the sum for each pair.
$$1-6=-5$$ $$2-3=-1$$
The solution is the pair that gives sum $-1$.
$$p=-3$$ $$q=2$$
Rewrite $a^{2}-a-6$ as $\left(a^{2}-3a\right)+\left(2a-6\right)$.
$$\left(a^{2}-3a\right)+\left(2a-6\right)$$
Factor out $a$ in the first and $2$ in the second group.
$$a\left(a-3\right)+2\left(a-3\right)$$
Factor out common term $a-3$ by using distributive property.
$$\left(a-3\right)\left(a+2\right)$$
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 $ax^{2}+bx+c=0$.
$$a^{2}-a-6=0$$
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.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$a^{2}-a-6=\left(a-3\right)\left(a+2\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -1x -6 = 0$$
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 $rs = C$
$$r + s = 1 $$ $$ rs = -6$$
Two numbers $r$ and $s$ sum up to $1$ exactly when the average of the two numbers is $\frac{1}{2}*1 = \frac{1}{2}$. 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 $u$.
$$r = \frac{1}{2} - u$$ $$s = \frac{1}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -6$
$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -6$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{1}{4} - u^2 = -6$$
Simplify the expression by subtracting $\frac{1}{4}$ on both sides
$$-u^2 = -6-\frac{1}{4} = -\frac{25}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
