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.
$$a+b=13$$ $$ab=-10\left(-3\right)=30$$
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 $13$.
$$a=10$$ $$b=3$$
Rewrite $-10x^{2}+13x-3$ as $\left(-10x^{2}+10x\right)+\left(3x-3\right)$.
$$\left(-10x^{2}+10x\right)+\left(3x-3\right)$$
Factor out $10x$ in the first and $-3$ in the second group.
$$10x\left(-x+1\right)-3\left(-x+1\right)$$
Factor out common term $-x+1$ by using distributive property.
