Step 1 :

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

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

$$ \begin{aligned} 4n^{3}+4n^{2}-9n-9 = ( \color{blue}{ 4x^{3}+4x^{2} } ) + ( \color{red}{ -9x-9 }) &= \\ &= \color{blue}{ 4x^{2}( x+1 )} + \color{red}{ -9( x+1 ) } = \\ &= (4x^{2}-9)(x+1) \end{aligned} $$

Step 2 :

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

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

Now we can apply the difference of squares formula.

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

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

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

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