Divide using synthetic division $$ ( 6x^{3}+29x^{2}+29x-14 ) \div ( 3x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-7&6&29&29&-14\\& & -42& 91& \color{black}{-840} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{120}&\color{orangered}{-854} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+29x^{2}+29x-14 }{ x+7 } = \color{blue}{6x^{2}-13x+120} \color{red}{~-~} \frac{ \color{red}{ 854 } }{ x+7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 7 = 0 $ ( $ x = \color{blue}{ -7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&6&29&29&-14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-7&\color{orangered}{ 6 }&29&29&-14\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 6 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&6&29&29&-14\\& & \color{blue}{-42} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-7&6&\color{orangered}{ 29 }&29&-14\\& & \color{orangered}{-42} & & \\ \hline &6&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 91 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&6&29&29&-14\\& & -42& \color{blue}{91} & \\ \hline &6&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ 91 } = \color{orangered}{ 120 } $
$$ \begin{array}{c|rrrr}-7&6&29&\color{orangered}{ 29 }&-14\\& & -42& \color{orangered}{91} & \\ \hline &6&-13&\color{orangered}{120}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 120 } = \color{blue}{ -840 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-7}&6&29&29&-14\\& & -42& 91& \color{blue}{-840} \\ \hline &6&-13&\color{blue}{120}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -840 \right) } = \color{orangered}{ -854 } $
$$ \begin{array}{c|rrrr}-7&6&29&29&\color{orangered}{ -14 }\\& & -42& 91& \color{orangered}{-840} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{120}&\color{orangered}{-854} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-13x+120 } $ with a remainder of $ \color{red}{ -854 } $.