Consider $6x^{2}-x-2$. Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+px+qx-2$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-1$$ $$pq=6\left(-2\right)=-12$$
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 $-12$.
$$1,-12$$ $$2,-6$$ $$3,-4$$
Calculate the sum for each pair.
$$1-12=-11$$ $$2-6=-4$$ $$3-4=-1$$
The solution is the pair that gives sum $-1$.
$$p=-4$$ $$q=3$$
Rewrite $6x^{2}-x-2$ as $\left(6x^{2}-4x\right)+\left(3x-2\right)$.
$$\left(6x^{2}-4x\right)+\left(3x-2\right)$$
Factor out $2x$ in $6x^{2}-4x$.
$$2x\left(3x-2\right)+3x-2$$
Factor out common term $3x-2$ by using distributive property.
