Factor polynomial $$ 16x^{3}-4x^{2}-36x+9 $$

$$ 16x^{3}-4x^{2}-36x+9 = ( 2x-3 )( 2x+3 )( 4x-1 ) $$

Step 1 :

To factor $ 16x^{3}-4x^{2}-36x+9 $ we can use factoring by grouping:

Group $ \color{blue}{ 16x^{3} }$ with $ \color{blue}{ -4x^{2} }$ and $ \color{red}{ -36x }$ with $ \color{red}{ 9 }$ then factor each group.

$$ \begin{aligned} 16x^{3}-4x^{2}-36x+9 = ( \color{blue}{ 16x^{3}-4x^{2} } ) + ( \color{red}{ -36x+9 }) &= \\ &= \color{blue}{ 4x^{2}( 4x-1 )} + \color{red}{ -9( 4x-1 ) } = \\ &= (4x^{2}-9)(4x-1) \end{aligned} $$

Step 2 :

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

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

Now we can apply the difference of squares formula.

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

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

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

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