Factor polynomial $$ 16y^{3}+32y^{2}-y-2 $$

$$ 16y^{3}+32y^{2}-y-2 = ( 4y-1 )( 4y+1 )( y+2 ) $$

Step 1 :

To factor $ 16y^{3}+32y^{2}-y-2 $ we can use factoring by grouping:

Group $ \color{blue}{ 16x^{3} }$ with $ \color{blue}{ 32x^{2} }$ and $ \color{red}{ -x }$ with $ \color{red}{ -2 }$ then factor each group.

$$ \begin{aligned} 16y^{3}+32y^{2}-y-2 = ( \color{blue}{ 16x^{3}+32x^{2} } ) + ( \color{red}{ -x-2 }) &= \\ &= \color{blue}{ 16x^{2}( x+2 )} + \color{red}{ -1( x+2 ) } = \\ &= (16x^{2}-1)(x+2) \end{aligned} $$

Step 2 :

Rewrite $ 16y^{2}-1 $ as:

$$ 16y^{2}-1 = (4y)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

After putting $ I = 4y $ and $ II = 1 $ , we have:

$$ 16y^{2}-1 = (4y)^2 - (1)^2 = ( 4y-1 ) ( 4y+1 ) $$

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