Factor polynomial $$ 8x^{4}+12x^{3}-32x^{2}-48x $$

$$ 8x^{4}+12x^{3}-32x^{2}-48x = 4x( x-2 )( x+2 )( 2x+3 ) $$

Step 1 :

After factoring out $ 4x $ we have:

$$ 8x^{4}+12x^{3}-32x^{2}-48x = 4x ( 2x^{3}+3x^{2}-8x-12 ) $$

Step 2 :

To factor $ 2x^{3}+3x^{2}-8x-12 $ we can use factoring by grouping:

Group $ \color{blue}{ 2x^{3} }$ with $ \color{blue}{ 3x^{2} }$ and $ \color{red}{ -8x }$ with $ \color{red}{ -12 }$ then factor each group.

$$ \begin{aligned} 2x^{3}+3x^{2}-8x-12 = ( \color{blue}{ 2x^{3}+3x^{2} } ) + ( \color{red}{ -8x-12 }) &= \\ &= \color{blue}{ x^{2}( 2x+3 )} + \color{red}{ -4( 2x+3 ) } = \\ &= (x^{2}-4)(2x+3) \end{aligned} $$

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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