Factor polynomial $$ x^{4}-x^{3}-4x^{2}+4x $$

$$ x^{4}-x^{3}-4x^{2}+4x = x( x-2 )( x+2 )( x-1 ) $$

Step 1 :

After factoring out $ x $ we have:

$$ x^{4}-x^{3}-4x^{2}+4x = x ( x^{3}-x^{2}-4x+4 ) $$

Step 2 :

To factor $ x^{3}-x^{2}-4x+4 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ -x^{2} }$ and $ \color{red}{ -4x }$ with $ \color{red}{ 4 }$ then factor each group.

$$ \begin{aligned} x^{3}-x^{2}-4x+4 = ( \color{blue}{ x^{3}-x^{2} } ) + ( \color{red}{ -4x+4 }) &= \\ &= \color{blue}{ x^{2}( x-1 )} + \color{red}{ -4( x-1 ) } = \\ &= (x^{2}-4)(x-1) \end{aligned} $$

Step 3 :

Rewrite $ x^{2}-4 $ as:

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

Now we can apply the difference of squares formula.

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

After putting $ I = x $ and $ II = 2 $ , we have:

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

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