Divide using synthetic division $$ ( 10x^{3}-14x^{2}-19x+4 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&10&-14&-19&4\\& & -20& 68& \color{black}{-98} \\ \hline &\color{blue}{10}&\color{blue}{-34}&\color{blue}{49}&\color{orangered}{-94} \end{array} $$The solution is:
$$ \frac{ 10x^{3}-14x^{2}-19x+4 }{ x+2 } = \color{blue}{10x^{2}-34x+49} \color{red}{~-~} \frac{ \color{red}{ 94 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&-14&-19&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 10 }&-14&-19&4\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&-14&-19&4\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -34 } $
$$ \begin{array}{c|rrrr}-2&10&\color{orangered}{ -14 }&-19&4\\& & \color{orangered}{-20} & & \\ \hline &10&\color{orangered}{-34}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -34 \right) } = \color{blue}{ 68 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&-14&-19&4\\& & -20& \color{blue}{68} & \\ \hline &10&\color{blue}{-34}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 68 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}-2&10&-14&\color{orangered}{ -19 }&4\\& & -20& \color{orangered}{68} & \\ \hline &10&-34&\color{orangered}{49}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 49 } = \color{blue}{ -98 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&10&-14&-19&4\\& & -20& 68& \color{blue}{-98} \\ \hline &10&-34&\color{blue}{49}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -98 \right) } = \color{orangered}{ -94 } $
$$ \begin{array}{c|rrrr}-2&10&-14&-19&\color{orangered}{ 4 }\\& & -20& 68& \color{orangered}{-98} \\ \hline &\color{blue}{10}&\color{blue}{-34}&\color{blue}{49}&\color{orangered}{-94} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-34x+49 } $ with a remainder of $ \color{red}{ -94 } $.