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.
$$\left(x+2\right)\left(x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+5x+6=0$$
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.
$$x=\frac{-5±\sqrt{5^{2}-4\times 6}}{2}$$
Square $5$.
$$x=\frac{-5±\sqrt{25-4\times 6}}{2}$$
Multiply $-4$ times $6$.
$$x=\frac{-5±\sqrt{25-24}}{2}$$
Add $25$ to $-24$.
$$x=\frac{-5±\sqrt{1}}{2}$$
Take the square root of $1$.
$$x=\frac{-5±1}{2}$$
Now solve the equation $x=\frac{-5±1}{2}$ when $±$ is plus. Add $-5$ to $1$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
Now solve the equation $x=\frac{-5±1}{2}$ when $±$ is minus. Subtract $1$ from $-5$.
$$x=-\frac{6}{2}$$
Divide $-6$ by $2$.
$$x=-3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-2$ for $x_{1}$ and $-3$ for $x_{2}$.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$x^{2}+5x+6=\left(x+2\right)\left(x+3\right)$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.
$$x ^ 2 +5x +6 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = -5 $$ $$ rs = 6$$
Two numbers $r$ and $s$ sum up to $-5$ exactly when the average of the two numbers is $\frac{1}{2}*-5 = -\frac{5}{2}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = -\frac{5}{2} - u$$ $$s = -\frac{5}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 6$
$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = 6$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{25}{4} - u^2 = 6$$
Simplify the expression by subtracting $\frac{25}{4}$ on both sides
$$-u^2 = 6-\frac{25}{4} = -\frac{1}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
