Consider $2x^{3}-3x^{2}-11x+6$. 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 positive, the positive number has greater absolute value than the negative. 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 $3$.
$$a=-1$$ $$b=4$$
Rewrite $2x^{2}+3x-2$ as $\left(2x^{2}-x\right)+\left(4x-2\right)$.
$$\left(2x^{2}-x\right)+\left(4x-2\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(2x-1\right)+2\left(2x-1\right)$$
Factor out common term $2x-1$ by using distributive property.
