Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=3\times 3=9$$
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 $9$.
$$-1,-9$$ $$-3,-3$$
Calculate the sum for each pair.
$$-1-9=-10$$ $$-3-3=-6$$
The solution is the pair that gives sum $-10$.
$$a=-9$$ $$b=-1$$
Rewrite $3x^{2}-10x+3$ as $\left(3x^{2}-9x\right)+\left(-x+3\right)$.
$$\left(3x^{2}-9x\right)+\left(-x+3\right)$$
Factor out $3x$ in the first and $-1$ in the second group.
$$3x\left(x-3\right)-\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(3x-1\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}-10x+3=0$$
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 $x=\frac{10±8}{6}$ when $±$ is plus. Add $10$ to $8$.
$$x=\frac{18}{6}$$
Divide $18$ by $6$.
$$x=3$$
Now solve the equation $x=\frac{10±8}{6}$ when $±$ is minus. Subtract $8$ from $10$.
$$x=\frac{2}{6}$$
Reduce the fraction $\frac{2}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{1}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $\frac{1}{3}$ for $x_{2}$.
