Divide using synthetic division $$ ( 3x^{3}+16x^{2}+19x-10 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&3&16&19&-10\\& & -9& -21& \color{black}{6} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+16x^{2}+19x-10 }{ x+3 } = \color{blue}{3x^{2}+7x-2} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&16&19&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 3 }&16&19&-10\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&16&19&-10\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-3&3&\color{orangered}{ 16 }&19&-10\\& & \color{orangered}{-9} & & \\ \hline &3&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&16&19&-10\\& & -9& \color{blue}{-21} & \\ \hline &3&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&3&16&\color{orangered}{ 19 }&-10\\& & -9& \color{orangered}{-21} & \\ \hline &3&7&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&16&19&-10\\& & -9& -21& \color{blue}{6} \\ \hline &3&7&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&3&16&19&\color{orangered}{ -10 }\\& & -9& -21& \color{orangered}{6} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+7x-2 } $ with a remainder of $ \color{red}{ -4 } $.