Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\times 2=2$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=2$$
Rewrite $x^{2}+3x+2$ as $\left(x^{2}+x\right)+\left(2x+2\right)$.
$$\left(x^{2}+x\right)+\left(2x+2\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x+1\right)+2\left(x+1\right)$$
Factor out common term $x+1$ by using distributive property.
$$\left(x+1\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+3x+2=0$$
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.
$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2}$$
Square $3$.
$$x=\frac{-3±\sqrt{9-4\times 2}}{2}$$
Multiply $-4$ times $2$.
$$x=\frac{-3±\sqrt{9-8}}{2}$$
Add $9$ to $-8$.
$$x=\frac{-3±\sqrt{1}}{2}$$
Take the square root of $1$.
$$x=\frac{-3±1}{2}$$
Now solve the equation $x=\frac{-3±1}{2}$ when $±$ is plus. Add $-3$ to $1$.
$$x=-\frac{2}{2}$$
Divide $-2$ by $2$.
$$x=-1$$
Now solve the equation $x=\frac{-3±1}{2}$ when $±$ is minus. Subtract $1$ from $-3$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-1$ for $x_{1}$ and $-2$ for $x_{2}$.
