Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=3\left(-10\right)=-30$$
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 $-30$.
$$-1,30$$ $$-2,15$$ $$-3,10$$ $$-5,6$$
Calculate the sum for each pair.
$$-1+30=29$$ $$-2+15=13$$ $$-3+10=7$$ $$-5+6=1$$
The solution is the pair that gives sum $1$.
$$a=-5$$ $$b=6$$
Rewrite $3x^{2}+x-10$ as $\left(3x^{2}-5x\right)+\left(6x-10\right)$.
$$\left(3x^{2}-5x\right)+\left(6x-10\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(3x-5\right)+2\left(3x-5\right)$$
Factor out common term $3x-5$ by using distributive property.
$$\left(3x-5\right)\left(x+2\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$3x^{2}+x-10=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Now solve the equation $x=\frac{-1±11}{6}$ when $±$ is plus. Add $-1$ to $11$.
$$x=\frac{10}{6}$$
Reduce the fraction $\frac{10}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{5}{3}$$
Now solve the equation $x=\frac{-1±11}{6}$ when $±$ is minus. Subtract $11$ from $-1$.
$$x=-\frac{12}{6}$$
Divide $-12$ by $6$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{3}$ for $x_{1}$ and $-2$ for $x_{2}$.
