Factor polynomial $$ 14x^{3}+12x^{2}-112x-96 $$
Answer
$$ 14x^{3}+12x^{2}-112x-96 = 2( x^{2}-8 )( 7x+6 ) $$
Explanation
Step 1 :
After factoring out $ 2 $ we have:
$$ 14x^{3}+12x^{2}-112x-96 = 2 ( 7x^{3}+6x^{2}-56x-48 ) $$Step 2 :
To factor $ 7x^{3}+6x^{2}-56x-48 $ we can use factoring by grouping:
Group $ \color{blue}{ 7x^{3} }$ with $ \color{blue}{ 6x^{2} }$ and $ \color{red}{ -56x }$ with $ \color{red}{ -48 }$ then factor each group.
$$ \begin{aligned} 7x^{3}+6x^{2}-56x-48 = ( \color{blue}{ 7x^{3}+6x^{2} } ) + ( \color{red}{ -56x-48 }) &= \\ &= \color{blue}{ x^{2}( 7x+6 )} + \color{red}{ -8( 7x+6 ) } = \\ &= (x^{2}-8)(7x+6) \end{aligned} $$This page was created using
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