Divide using synthetic division $$ ( 3x^{4}+19x^{3}+16x ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&19&0&16&0\\& & -6& -26& 52& \color{black}{-136} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{-26}&\color{blue}{68}&\color{orangered}{-136} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+19x^{3}+16x }{ x+2 } = \color{blue}{3x^{3}+13x^{2}-26x+68} \color{red}{~-~} \frac{ \color{red}{ 136 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&19&0&16&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&19&0&16&0\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&19&0&16&0\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 19 }&0&16&0\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 13 } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&19&0&16&0\\& & -6& \color{blue}{-26} & & \\ \hline &3&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrrr}-2&3&19&\color{orangered}{ 0 }&16&0\\& & -6& \color{orangered}{-26} & & \\ \hline &3&13&\color{orangered}{-26}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ 52 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&19&0&16&0\\& & -6& -26& \color{blue}{52} & \\ \hline &3&13&\color{blue}{-26}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 52 } = \color{orangered}{ 68 } $
$$ \begin{array}{c|rrrrr}-2&3&19&0&\color{orangered}{ 16 }&0\\& & -6& -26& \color{orangered}{52} & \\ \hline &3&13&-26&\color{orangered}{68}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 68 } = \color{blue}{ -136 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&19&0&16&0\\& & -6& -26& 52& \color{blue}{-136} \\ \hline &3&13&-26&\color{blue}{68}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -136 \right) } = \color{orangered}{ -136 } $
$$ \begin{array}{c|rrrrr}-2&3&19&0&16&\color{orangered}{ 0 }\\& & -6& -26& 52& \color{orangered}{-136} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{-26}&\color{blue}{68}&\color{orangered}{-136} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+13x^{2}-26x+68 } $ with a remainder of $ \color{red}{ -136 } $.