By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-6$ and $q$ divides the leading coefficient $2$. One such root is $-3$. Factor the polynomial by dividing it by $x+3$.
$$\left(x+3\right)\left(2x^{2}-3x-2\right)$$
Consider $2x^{2}-3x-2$. Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-3$$ $$ab=2\left(-2\right)=-4$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-4$.
$$1,-4$$ $$2,-2$$
Calculate the sum for each pair.
$$1-4=-3$$ $$2-2=0$$
The solution is the pair that gives sum $-3$.
$$a=-4$$ $$b=1$$
Rewrite $2x^{2}-3x-2$ as $\left(2x^{2}-4x\right)+\left(x-2\right)$.
$$\left(2x^{2}-4x\right)+\left(x-2\right)$$
Factor out $2x$ in $2x^{2}-4x$.
$$2x\left(x-2\right)+x-2$$
Factor out common term $x-2$ by using distributive property.
