Divide using synthetic division $$ ( 10x^{3}+19x-10 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&10&0&19&-10\\& & -20& 40& \color{black}{-118} \\ \hline &\color{blue}{10}&\color{blue}{-20}&\color{blue}{59}&\color{orangered}{-128} \end{array} $$The solution is:
$$ \frac{ 10x^{3}+19x-10 }{ x+2 } = \color{blue}{10x^{2}-20x+59} \color{red}{~-~} \frac{ \color{red}{ 128 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&0&19&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 10 }&0&19&-10\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&0&19&-10\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}-2&10&\color{orangered}{ 0 }&19&-10\\& & \color{orangered}{-20} & & \\ \hline &10&\color{orangered}{-20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&0&19&-10\\& & -20& \color{blue}{40} & \\ \hline &10&\color{blue}{-20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 40 } = \color{orangered}{ 59 } $
$$ \begin{array}{c|rrrr}-2&10&0&\color{orangered}{ 19 }&-10\\& & -20& \color{orangered}{40} & \\ \hline &10&-20&\color{orangered}{59}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 59 } = \color{blue}{ -118 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&0&19&-10\\& & -20& 40& \color{blue}{-118} \\ \hline &10&-20&\color{blue}{59}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -118 \right) } = \color{orangered}{ -128 } $
$$ \begin{array}{c|rrrr}-2&10&0&19&\color{orangered}{ -10 }\\& & -20& 40& \color{orangered}{-118} \\ \hline &\color{blue}{10}&\color{blue}{-20}&\color{blue}{59}&\color{orangered}{-128} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-20x+59 } $ with a remainder of $ \color{red}{ -128 } $.