By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-2$ and $q$ divides the leading coefficient $6$. One such root is $-1$. Factor the polynomial by dividing it by $x+1$.
$$\left(x+1\right)\left(6x^{2}+x-2\right)$$
Consider $6x^{2}+x-2$. Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=6\left(-2\right)=-12$$
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 $-12$.
$$-1,12$$ $$-2,6$$ $$-3,4$$
Calculate the sum for each pair.
$$-1+12=11$$ $$-2+6=4$$ $$-3+4=1$$
The solution is the pair that gives sum $1$.
$$a=-3$$ $$b=4$$
Rewrite $6x^{2}+x-2$ as $\left(6x^{2}-3x\right)+\left(4x-2\right)$.
$$\left(6x^{2}-3x\right)+\left(4x-2\right)$$
Factor out $3x$ in the first and $2$ in the second group.
$$3x\left(2x-1\right)+2\left(2x-1\right)$$
Factor out common term $2x-1$ by using distributive property.
