Question
Factor polynomial $$ 96n^{3}-84n^{2}+112n-98 $$
Answer
$$ 96n^{3}-84n^{2}+112n-98 = 2( 6n^{2}+7 )( 8n-7 ) $$
Explanation
Step 1 :
After factoring out $ 2 $ we have:
$$ 96n^{3}-84n^{2}+112n-98 = 2 ( 48n^{3}-42n^{2}+56n-49 ) $$Step 2 :
To factor $ 48n^{3}-42n^{2}+56n-49 $ we can use factoring by grouping:
Group $ \color{blue}{ 48x^{3} }$ with $ \color{blue}{ -42x^{2} }$ and $ \color{red}{ 56x }$ with $ \color{red}{ -49 }$ then factor each group.
$$ \begin{aligned} 48n^{3}-42n^{2}+56n-49 = ( \color{blue}{ 48x^{3}-42x^{2} } ) + ( \color{red}{ 56x-49 }) &= \\ &= \color{blue}{ 6x^{2}( 8x-7 )} + \color{red}{ 7( 8x-7 ) } = \\ &= (6x^{2}+7)(8x-7) \end{aligned} $$This page was created using
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