By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-30$ and $q$ divides the leading coefficient $2$. One such root is $-3$. Factor the polynomial by dividing it by $x+3$.
$$\left(x+3\right)\left(2x^{2}+x-10\right)$$
Consider $2x^{2}+x-10$. Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=2\left(-10\right)=-20$$
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 $-20$.
$$-1,20$$ $$-2,10$$ $$-4,5$$
Calculate the sum for each pair.
$$-1+20=19$$ $$-2+10=8$$ $$-4+5=1$$
The solution is the pair that gives sum $1$.
$$a=-4$$ $$b=5$$
Rewrite $2x^{2}+x-10$ as $\left(2x^{2}-4x\right)+\left(5x-10\right)$.
$$\left(2x^{2}-4x\right)+\left(5x-10\right)$$
Factor out $2x$ in the first and $5$ in the second group.
$$2x\left(x-2\right)+5\left(x-2\right)$$
Factor out common term $x-2$ by using distributive property.
