Consider $x^{2}+5x+6$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=1\times 6=6$$
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 $6$.
$$1,6$$ $$2,3$$
Calculate the sum for each pair.
$$1+6=7$$ $$2+3=5$$
The solution is the pair that gives sum $5$.
$$a=2$$ $$b=3$$
Rewrite $x^{2}+5x+6$ as $\left(x^{2}+2x\right)+\left(3x+6\right)$.
$$\left(x^{2}+2x\right)+\left(3x+6\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(x+2\right)+3\left(x+2\right)$$
Factor out common term $x+2$ by using distributive property.
$$\left(x+2\right)\left(x+3\right)$$
Rewrite the complete factored expression.
$$2\left(x+2\right)\left(x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}+10x+12=0$$
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.
