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

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Answer 1 · 2015-08-12T06:29:00

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

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

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 $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 1*4 = 4$
AND
$N_1 +N_2 = b = 4$

After trying out a few numbers we get $N_1 = 2$ and $N_2 =2$
$2*2 = 4$ , and $2+2= 4$

$x^2+color(blue)(4x)+4= x^2+color(blue)(2x +2x)+4=0$

$=x(x+2 ) +2(x+2)$

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

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

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

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