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

$$ x^{3}+x^{2}-9x-9 = ( x-3 )( x+3 )( x+1 ) $$

Step 1 :

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

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

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

Step 2 :

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

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

Now we can apply the difference of squares formula.

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

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

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

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