Factor polynomial $$ 9x^{3}+18x^{2}-4x-8 $$

$$ 9x^{3}+18x^{2}-4x-8 = ( 3x-2 )( 3x+2 )( x+2 ) $$

Step 1 :

To factor $ 9x^{3}+18x^{2}-4x-8 $ we can use factoring by grouping:

Group $ \color{blue}{ 9x^{3} }$ with $ \color{blue}{ 18x^{2} }$ and $ \color{red}{ -4x }$ with $ \color{red}{ -8 }$ then factor each group.

$$ \begin{aligned} 9x^{3}+18x^{2}-4x-8 = ( \color{blue}{ 9x^{3}+18x^{2} } ) + ( \color{red}{ -4x-8 }) &= \\ &= \color{blue}{ 9x^{2}( x+2 )} + \color{red}{ -4( x+2 ) } = \\ &= (9x^{2}-4)(x+2) \end{aligned} $$

Step 2 :

Rewrite $ 9x^{2}-4 $ as:

$$ 9x^{2}-4 = (3x)^2 - (2)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = 3x $ and $ II = 2 $ , we have:

$$ 9x^{2}-4 = (3x)^2 - (2)^2 = ( 3x-2 ) ( 3x+2 ) $$

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