# How do you solve x^2+4x+4=0 by factoring?

Aug 12, 2015

The solution is
color(blue)(x=-2

#### Explanation:

${x}^{2} + 4 x + 4 = 0$

We can Split the Middle Term of this expression to factorise it and thereby find the solutions:

In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot 4 = 4$
AND
${N}_{1} + {N}_{2} = b = 4$

After trying out a few numbers we get ${N}_{1} = 2$ and ${N}_{2} = 2$
$2 \cdot 2 = 4$, and $2 + 2 = 4$

${x}^{2} + \textcolor{b l u e}{4 x} + 4 = {x}^{2} + \textcolor{b l u e}{2 x + 2 x} + 4 = 0$

$= x \left(x + 2\right) + 2 \left(x + 2\right)$

 color(blue)((x+2)(x+2 ) are the factors of the expression.

Now we equate these factors to zero to find the solutions.

 x+2 = 0 , color(blue)(x=-2
x+2 = 0, color(blue)(x=-2