Question
Factor polynomial $$ 3u^{3}+4u^{2}+9u+12 $$
Answer
$$ 3u^{3}+4u^{2}+9u+12 = ( u^{2}+3 )( 3u+4 ) $$
Explanation
To factor $ 3u^{3}+4u^{2}+9u+12 $ we can use factoring by grouping:
Group $ \color{blue}{ 3x^{3} }$ with $ \color{blue}{ 4x^{2} }$ and $ \color{red}{ 9x }$ with $ \color{red}{ 12 }$ then factor each group.
$$ \begin{aligned} 3u^{3}+4u^{2}+9u+12 = ( \color{blue}{ 3x^{3}+4x^{2} } ) + ( \color{red}{ 9x+12 }) &= \\ &= \color{blue}{ x^{2}( 3x+4 )} + \color{red}{ 3( 3x+4 ) } = \\ &= (x^{2}+3)(3x+4) \end{aligned} $$This page was created using
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