Factor polynomial $$ 4x^{3}-24x^{2}-16x+96 $$

$$ 4x^{3}-24x^{2}-16x+96 = 4( x-2 )( x+2 )( x-6 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 4x^{3}-24x^{2}-16x+96 = 4 ( x^{3}-6x^{2}-4x+24 ) $$

Step 2 :

To factor $ x^{3}-6x^{2}-4x+24 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ -6x^{2} }$ and $ \color{red}{ -4x }$ with $ \color{red}{ 24 }$ then factor each group.

$$ \begin{aligned} x^{3}-6x^{2}-4x+24 = ( \color{blue}{ x^{3}-6x^{2} } ) + ( \color{red}{ -4x+24 }) &= \\ &= \color{blue}{ x^{2}( x-6 )} + \color{red}{ -4( x-6 ) } = \\ &= (x^{2}-4)(x-6) \end{aligned} $$

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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