Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-5$$ $$ab=2\left(-3\right)=-6$$
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 $-6$.
$$1,-6$$ $$2,-3$$
Calculate the sum for each pair.
$$1-6=-5$$ $$2-3=-1$$
The solution is the pair that gives sum $-5$.
$$a=-6$$ $$b=1$$
Rewrite $2x^{2}-5x-3$ as $\left(2x^{2}-6x\right)+\left(x-3\right)$.
$$\left(2x^{2}-6x\right)+\left(x-3\right)$$
Factor out $2x$ in $2x^{2}-6x$.
$$2x\left(x-3\right)+x-3$$
Factor out common term $x-3$ by using distributive property.
