Factor polynomial $$ 8x^{3}+2x^{2}-8x-2 $$

$$ 8x^{3}+2x^{2}-8x-2 = 2( x-1 )( x+1 )( 4x+1 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 8x^{3}+2x^{2}-8x-2 = 2 ( 4x^{3}+x^{2}-4x-1 ) $$

Step 2 :

To factor $ 4x^{3}+x^{2}-4x-1 $ we can use factoring by grouping:

Group $ \color{blue}{ 4x^{3} }$ with $ \color{blue}{ x^{2} }$ and $ \color{red}{ -4x }$ with $ \color{red}{ -1 }$ then factor each group.

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

Step 3 :

Rewrite $ x^{2}-1 $ as:

$$ x^{2}-1 = (x)^2 - (1)^2 $$

Now we can apply the difference of squares formula.

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

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

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

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