Factor polynomial $$ 2x^{2}-8x+8 $$

$$ 2x^{2}-8x+8 = 2( x-2 )^{ 2 } $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2x^{2}-8x+8 = 2 ( x^{2}-4x+4 ) $$

Step 2 :

Both the first and third terms are perfect squares.

$$ x^2 = \left( \color{blue}{ x } \right)^2 ~~ \text{and} ~~ 4 = \left( \color{red}{ 2 } \right)^2 $$

The middle term ( $ -4x $ ) is two times the product of the terms that are squared.

$$ -4x = - 2 \cdot \color{blue}{x} \cdot \color{red}{2} $$

We can conclude that the polynomial $ x^{2}-4x+4 $ is a perfect square trinomial, so we will use the formula below.

$$ A^2 - 2AB + B^2 = (A - B)^2 $$

In this example we have $ \color{blue}{ A = x } $ and $ \color{red}{ B = 2 } $ so,

$$ x^{2}-4x+4 = ( \color{blue}{ x } - \color{red}{ 2 } )^2 $$

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