Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=5$$ $$ab=1\times 4=4$$
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 $4$.
$$1,4$$ $$2,2$$
Calculate the sum for each pair.
$$1+4=5$$ $$2+2=4$$
The solution is the pair that gives sum $5$.
$$a=1$$ $$b=4$$
Rewrite $x^{2}+5x+4$ as $\left(x^{2}+x\right)+\left(4x+4\right)$.
$$\left(x^{2}+x\right)+\left(4x+4\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(x+1\right)+4\left(x+1\right)$$
Factor out common term $x+1$ by using distributive property.
$$\left(x+1\right)\left(x+4\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+5x+4=0$$
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.
$$x=\frac{-5±\sqrt{5^{2}-4\times 4}}{2}$$
Square $5$.
$$x=\frac{-5±\sqrt{25-4\times 4}}{2}$$
Multiply $-4$ times $4$.
$$x=\frac{-5±\sqrt{25-16}}{2}$$
Add $25$ to $-16$.
$$x=\frac{-5±\sqrt{9}}{2}$$
Take the square root of $9$.
$$x=\frac{-5±3}{2}$$
Now solve the equation $x=\frac{-5±3}{2}$ when $±$ is plus. Add $-5$ to $3$.
$$x=-\frac{2}{2}$$
Divide $-2$ by $2$.
$$x=-1$$
Now solve the equation $x=\frac{-5±3}{2}$ when $±$ is minus. Subtract $3$ from $-5$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
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 $-4$ for $x_{2}$.
