Factor polynomial $$ 3k^{3}-4k^{2}-12k+16 $$

$$ 3k^{3}-4k^{2}-12k+16 = ( k-2 )( k+2 )( 3k-4 ) $$

Step 1 :

To factor $ 3k^{3}-4k^{2}-12k+16 $ we can use factoring by grouping:

Group $ \color{blue}{ 3x^{3} }$ with $ \color{blue}{ -4x^{2} }$ and $ \color{red}{ -12x }$ with $ \color{red}{ 16 }$ then factor each group.

$$ \begin{aligned} 3k^{3}-4k^{2}-12k+16 = ( \color{blue}{ 3x^{3}-4x^{2} } ) + ( \color{red}{ -12x+16 }) &= \\ &= \color{blue}{ x^{2}( 3x-4 )} + \color{red}{ -4( 3x-4 ) } = \\ &= (x^{2}-4)(3x-4) \end{aligned} $$

Step 2 :

Rewrite $ k^{2}-4 $ as:

$$ k^{2}-4 = (k)^2 - (2)^2 $$

Now we can apply the difference of squares formula.

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

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

$$ k^{2}-4 = (k)^2 - (2)^2 = ( k-2 ) ( k+2 ) $$

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