Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+2\right)^{2}$.
$$x^{2}+4x+4=0$$
To solve the equation, factor $x^{2}+4x+4$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=4$$
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 $4$.
$$a=2$$ $$b=2$$
Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x+2\right)\left(x+2\right)$$
Rewrite as a binomial square.
$$\left(x+2\right)^{2}$$
To find equation solution, solve $x+2=0$.
$$x=-2$$
Steps Using Factoring By Grouping
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+2\right)^{2}$.
$$x^{2}+4x+4=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=1\times 4=4$$
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 $4$.
$$a=2$$ $$b=2$$
Rewrite $x^{2}+4x+4$ as $\left(x^{2}+2x\right)+\left(2x+4\right)$.
$$\left(x^{2}+2x\right)+\left(2x+4\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x+2\right)+2\left(x+2\right)$$
Factor out common term $x+2$ by using distributive property.
$$\left(x+2\right)\left(x+2\right)$$
Rewrite as a binomial square.
$$\left(x+2\right)^{2}$$
To find equation solution, solve $x+2=0$.
$$x=-2$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+2\right)^{2}$.
$$x^{2}+4x+4=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $4$ for $b$, and $4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-4±\sqrt{4^{2}-4\times 4}}{2}$$
Square $4$.
$$x=\frac{-4±\sqrt{16-4\times 4}}{2}$$
Multiply $-4$ times $4$.
$$x=\frac{-4±\sqrt{16-16}}{2}$$
Add $16$ to $-16$.
$$x=\frac{-4±\sqrt{0}}{2}$$
Take the square root of $0$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
Steps for Completing the Square
Take the square root of both sides of the equation.
$$\sqrt{\left(x+2\right)^{2}}=\sqrt{0}$$
Simplify.
$$x+2=0$$ $$x+2=0$$
Subtract $2$ from both sides of the equation.
$$x=-2$$ $$x=-2$$
The equation is now solved. Solutions are the same.
