Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=1\left(-6\right)=-6$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. 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 $1$.
$$a=-2$$ $$b=3$$
Rewrite $x^{2}+x-6$ as $\left(x^{2}-2x\right)+\left(3x-6\right)$.
$$\left(x^{2}-2x\right)+\left(3x-6\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(x-2\right)+3\left(x-2\right)$$
Factor out common term $x-2$ by using distributive property.
$$\left(x-2\right)\left(x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+x-6=0$$
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.
$$x=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2}$$
Square $1$.
$$x=\frac{-1±\sqrt{1-4\left(-6\right)}}{2}$$
Multiply $-4$ times $-6$.
$$x=\frac{-1±\sqrt{1+24}}{2}$$
Add $1$ to $24$.
$$x=\frac{-1±\sqrt{25}}{2}$$
Take the square root of $25$.
$$x=\frac{-1±5}{2}$$
Now solve the equation $x=\frac{-1±5}{2}$ when $±$ is plus. Add $-1$ to $5$.
$$x=\frac{4}{2}$$
Divide $4$ by $2$.
$$x=2$$
Now solve the equation $x=\frac{-1±5}{2}$ when $±$ is minus. Subtract $5$ from $-1$.
$$x=-\frac{6}{2}$$
Divide $-6$ by $2$.
$$x=-3$$
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 $-3$ for $x_{2}$.
