Factor polynomial $$ -2x^{4}+4x^{2}-2 $$

Answer

$$ -2x^{4}+4x^{2}-2 = ( -2 )( x-1 )^{ 2 }( x+1 )^{ 2 } $$

Step 1 :

After factoring out $ -2 $ we have:

$$ -2x^{4}+4x^{2}-2 = -2 ( x^{4}-2x^{2}+1 ) $$

Step 2 :

Both the first and third terms are perfect squares.

$$ x^4 = \left( \color{blue}{ x^{2} } \right)^2 ~~ \text{and} ~~ 1 = \left( \color{red}{ 1 } \right)^2 $$

The middle term ( $ -2x^2 $ ) is two times the product of the terms that are squared.

$$ -2x^2 = - 2 \cdot \color{blue}{x^{2}} \cdot \color{red}{1} $$

We can conclude that the polynomial $ x^{4}-2x^{2}+1 $ 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^{2} } $ and $ \color{red}{ B = 1 } $ so,

$$ x^{4}-2x^{2}+1 = ( \color{blue}{ x^{2} } - \color{red}{ 1 } )^2 $$

Step 3 :

Rewrite $ x^{2}-1 $ as:

$$ x^{2}-1 = (x)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

$$ I^2 - II^2 = (I - II)(I + II) $$

After putting $ I = x $ and $ II = 1 $ , we have:

$$ x^{2}-1 = (x)^2 - (1)^2 = ( x-1 ) ( x+1 ) $$

Step 4 :