Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+28$. To find $a$ and $b$, set up a system to be solved.
$$a+b=11$$ $$ab=1\times 28=28$$
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 $28$.
$$1,28$$ $$2,14$$ $$4,7$$
Calculate the sum for each pair.
$$1+28=29$$ $$2+14=16$$ $$4+7=11$$
The solution is the pair that gives sum $11$.
$$a=4$$ $$b=7$$
Rewrite $x^{2}+11x+28$ as $\left(x^{2}+4x\right)+\left(7x+28\right)$.
$$\left(x^{2}+4x\right)+\left(7x+28\right)$$
Factor out $x$ in the first and $7$ in the second group.
$$x\left(x+4\right)+7\left(x+4\right)$$
Factor out common term $x+4$ by using distributive property.
$$\left(x+4\right)\left(x+7\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+11x+28=0$$
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.
$$x=\frac{-11±\sqrt{11^{2}-4\times 28}}{2}$$
Square $11$.
$$x=\frac{-11±\sqrt{121-4\times 28}}{2}$$
Multiply $-4$ times $28$.
$$x=\frac{-11±\sqrt{121-112}}{2}$$
Add $121$ to $-112$.
$$x=\frac{-11±\sqrt{9}}{2}$$
Take the square root of $9$.
$$x=\frac{-11±3}{2}$$
Now solve the equation $x=\frac{-11±3}{2}$ when $±$ is plus. Add $-11$ to $3$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
Now solve the equation $x=\frac{-11±3}{2}$ when $±$ is minus. Subtract $3$ from $-11$.
$$x=-\frac{14}{2}$$
Divide $-14$ by $2$.
$$x=-7$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-4$ for $x_{1}$ and $-7$ for $x_{2}$.
