Factor polynomial $$ c^{3}+3c^{2}-81c-243 $$

$$ c^{3}+3c^{2}-81c-243 = ( c-9 )( c+9 )( c+3 ) $$

Step 1 :

To factor $ c^{3}+3c^{2}-81c-243 $ we can use factoring by grouping:

Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ 3x^{2} }$ and $ \color{red}{ -81x }$ with $ \color{red}{ -243 }$ then factor each group.

$$ \begin{aligned} c^{3}+3c^{2}-81c-243 = ( \color{blue}{ x^{3}+3x^{2} } ) + ( \color{red}{ -81x-243 }) &= \\ &= \color{blue}{ x^{2}( x+3 )} + \color{red}{ -81( x+3 ) } = \\ &= (x^{2}-81)(x+3) \end{aligned} $$

Step 2 :

Rewrite $ c^{2}-81 $ as:

$$ c^{2}-81 = (c)^2 - (9)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = c $ and $ II = 9 $ , we have:

$$ c^{2}-81 = (c)^2 - (9)^2 = ( c-9 ) ( c+9 ) $$

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