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
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.
$$3x^{2}+12x+12=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $12$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$3x^{2}+12x+12=0$$
Subtract $12$ from both sides of the equation.
$$3x^{2}+12x+12-12=-12$$
Subtracting $12$ from itself leaves $0$.
$$3x^{2}+12x=-12$$
Divide both sides by $3$.
$$\frac{3x^{2}+12x}{3}=-\frac{12}{3}$$
Dividing by $3$ undoes the multiplication by $3$.
$$x^{2}+\frac{12}{3}x=-\frac{12}{3}$$
Divide $12$ by $3$.
$$x^{2}+4x=-\frac{12}{3}$$
Divide $-12$ by $3$.
$$x^{2}+4x=-4$$
Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+4x+2^{2}=-4+2^{2}$$
Square $2$.
$$x^{2}+4x+4=-4+4$$
Add $-4$ to $4$.
$$x^{2}+4x+4=0$$
Factor $x^{2}+4x+4$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+2\right)^{2}=0$$
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.
$$x=-2$$
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 $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $3$
$$x ^ 2 +4x +4 = 0$$
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 $rs = C$
$$r + s = -4 $$ $$ rs = 4$$
Two numbers $r$ and $s$ sum up to $-4$ exactly when the average of the two numbers is $\frac{1}{2}*-4 = -2$. 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 $u$.
$$r = -2 - u$$ $$s = -2 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 4$
$$(-2 - u) (-2 + u) = 4$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$4 - u^2 = 4$$
Simplify the expression by subtracting $4$ on both sides
$$-u^2 = 4-4 = 0$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 0$$ $$u = 0 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
