Divide using synthetic division $$ ( 6x^{4}+21x^{3}+10x^{2}-29 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&6&21&10&0&-29\\& & -18& -9& -3& \color{black}{9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+21x^{3}+10x^{2}-29 }{ x+3 } = \color{blue}{6x^{3}+3x^{2}+x-3} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&10&0&-29\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 6 }&21&10&0&-29\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&10&0&-29\\& & \color{blue}{-18} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&6&\color{orangered}{ 21 }&10&0&-29\\& & \color{orangered}{-18} & & & \\ \hline &6&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&10&0&-29\\& & -18& \color{blue}{-9} & & \\ \hline &6&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&6&21&\color{orangered}{ 10 }&0&-29\\& & -18& \color{orangered}{-9} & & \\ \hline &6&3&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&10&0&-29\\& & -18& -9& \color{blue}{-3} & \\ \hline &6&3&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&6&21&10&\color{orangered}{ 0 }&-29\\& & -18& -9& \color{orangered}{-3} & \\ \hline &6&3&1&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&10&0&-29\\& & -18& -9& -3& \color{blue}{9} \\ \hline &6&3&1&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 9 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-3&6&21&10&0&\color{orangered}{ -29 }\\& & -18& -9& -3& \color{orangered}{9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+3x^{2}+x-3 } $ with a remainder of $ \color{red}{ -20 } $.