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 $x+3$.
$$\left(x+3\right)\left(2x^{2}-9x+10\right)$$
Consider $2x^{2}-9x+10$. 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.
$$a+b=-9$$ $$ab=2\times 10=20$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $20$.
$$-1,-20$$ $$-2,-10$$ $$-4,-5$$
Calculate the sum for each pair.
$$-1-20=-21$$ $$-2-10=-12$$ $$-4-5=-9$$
The solution is the pair that gives sum $-9$.
$$a=-5$$ $$b=-4$$
Rewrite $2x^{2}-9x+10$ as $\left(2x^{2}-5x\right)+\left(-4x+10\right)$.
$$\left(2x^{2}-5x\right)+\left(-4x+10\right)$$
Factor out $x$ in the first and $-2$ in the second group.
$$x\left(2x-5\right)-2\left(2x-5\right)$$
Factor out common term $2x-5$ by using distributive property.
