Factor the expression by grouping. First, the expression needs to be rewritten as $10x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=11$$ $$ab=10\times 3=30$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $30$.
$$1,30$$ $$2,15$$ $$3,10$$ $$5,6$$
Calculate the sum for each pair.
$$1+30=31$$ $$2+15=17$$ $$3+10=13$$ $$5+6=11$$
The solution is the pair that gives sum $11$.
$$a=5$$ $$b=6$$
Rewrite $10x^{2}+11x+3$ as $\left(10x^{2}+5x\right)+\left(6x+3\right)$.
$$\left(10x^{2}+5x\right)+\left(6x+3\right)$$
Factor out $5x$ in the first and $3$ in the second group.
$$5x\left(2x+1\right)+3\left(2x+1\right)$$
Factor out common term $2x+1$ by using distributive property.
