Divide using synthetic division $$ ( 2x^{4}+6x^{3}-10x^{2}-13x-29 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&6&-10&-13&-29\\& & -8& 8& 8& \color{black}{20} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-5}&\color{orangered}{-9} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+6x^{3}-10x^{2}-13x-29 }{ x+4 } = \color{blue}{2x^{3}-2x^{2}-2x-5} \color{red}{~-~} \frac{ \color{red}{ 9 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&6&-10&-13&-29\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&6&-10&-13&-29\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&6&-10&-13&-29\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 6 }&-10&-13&-29\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&6&-10&-13&-29\\& & -8& \color{blue}{8} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 8 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-4&2&6&\color{orangered}{ -10 }&-13&-29\\& & -8& \color{orangered}{8} & & \\ \hline &2&-2&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&6&-10&-13&-29\\& & -8& 8& \color{blue}{8} & \\ \hline &2&-2&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 8 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-4&2&6&-10&\color{orangered}{ -13 }&-29\\& & -8& 8& \color{orangered}{8} & \\ \hline &2&-2&-2&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&6&-10&-13&-29\\& & -8& 8& 8& \color{blue}{20} \\ \hline &2&-2&-2&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 20 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-4&2&6&-10&-13&\color{orangered}{ -29 }\\& & -8& 8& 8& \color{orangered}{20} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-5}&\color{orangered}{-9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x^{2}-2x-5 } $ with a remainder of $ \color{red}{ -9 } $.