Step 1 :

To factor $ 4m^{3}+9m^{2}-36m-81 $ we can use factoring by grouping:

Group $ \color{blue}{ 4x^{3} }$ with $ \color{blue}{ 9x^{2} }$ and $ \color{red}{ -36x }$ with $ \color{red}{ -81 }$ then factor each group.

$$ \begin{aligned} 4m^{3}+9m^{2}-36m-81 = ( \color{blue}{ 4x^{3}+9x^{2} } ) + ( \color{red}{ -36x-81 }) &= \\ &= \color{blue}{ x^{2}( 4x+9 )} + \color{red}{ -9( 4x+9 ) } = \\ &= (x^{2}-9)(4x+9) \end{aligned} $$

Step 2 :

Rewrite $ m^{2}-9 $ as:

$$ m^{2}-9 = (m)^2 - (3)^2 $$

Now we can apply the difference of squares formula.

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

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

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

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