Factor polynomial $$ 42b^{3}+30b^{2}+98b+70 $$
Answer
$$ 42b^{3}+30b^{2}+98b+70 = 2( 3b^{2}+7 )( 7b+5 ) $$
Explanation
Step 1 :
After factoring out $ 2 $ we have:
$$ 42b^{3}+30b^{2}+98b+70 = 2 ( 21b^{3}+15b^{2}+49b+35 ) $$Step 2 :
To factor $ 21b^{3}+15b^{2}+49b+35 $ we can use factoring by grouping:
Group $ \color{blue}{ 21x^{3} }$ with $ \color{blue}{ 15x^{2} }$ and $ \color{red}{ 49x }$ with $ \color{red}{ 35 }$ then factor each group.
$$ \begin{aligned} 21b^{3}+15b^{2}+49b+35 = ( \color{blue}{ 21x^{3}+15x^{2} } ) + ( \color{red}{ 49x+35 }) &= \\ &= \color{blue}{ 3x^{2}( 7x+5 )} + \color{red}{ 7( 7x+5 ) } = \\ &= (3x^{2}+7)(7x+5) \end{aligned} $$This page was created using
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