Factor polynomial $$ y^{3}+4y^{2}+7y+28 $$
Answer
$$ y^{3}+4y^{2}+7y+28 = ( y^{2}+7 )( y+4 ) $$
Explanation
To factor $ y^{3}+4y^{2}+7y+28 $ we can use factoring by grouping:
Group $ \color{blue}{ x^{3} }$ with $ \color{blue}{ 4x^{2} }$ and $ \color{red}{ 7x }$ with $ \color{red}{ 28 }$ then factor each group.
$$ \begin{aligned} y^{3}+4y^{2}+7y+28 = ( \color{blue}{ x^{3}+4x^{2} } ) + ( \color{red}{ 7x+28 }) &= \\ &= \color{blue}{ x^{2}( x+4 )} + \color{red}{ 7( x+4 ) } = \\ &= (x^{2}+7)(x+4) \end{aligned} $$This page was created using
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