Divide using synthetic division $$ ( 2x^{4}+17x^{3}-29x^{2}+21 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-10&2&17&-29&0&21\\& & -20& 30& -10& \color{black}{100} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{1}&\color{blue}{-10}&\color{orangered}{121} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+17x^{3}-29x^{2}+21 }{ x+10 } = \color{blue}{2x^{3}-3x^{2}+x-10} ~+~ \frac{ \color{red}{ 121 } }{ x+10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&2&17&-29&0&21\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-10&\color{orangered}{ 2 }&17&-29&0&21\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 2 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&2&17&-29&0&21\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-10&2&\color{orangered}{ 17 }&-29&0&21\\& & \color{orangered}{-20} & & & \\ \hline &2&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&2&17&-29&0&21\\& & -20& \color{blue}{30} & & \\ \hline &2&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 30 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-10&2&17&\color{orangered}{ -29 }&0&21\\& & -20& \color{orangered}{30} & & \\ \hline &2&-3&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&2&17&-29&0&21\\& & -20& 30& \color{blue}{-10} & \\ \hline &2&-3&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-10&2&17&-29&\color{orangered}{ 0 }&21\\& & -20& 30& \color{orangered}{-10} & \\ \hline &2&-3&1&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-10}&2&17&-29&0&21\\& & -20& 30& -10& \color{blue}{100} \\ \hline &2&-3&1&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 100 } = \color{orangered}{ 121 } $
$$ \begin{array}{c|rrrrr}-10&2&17&-29&0&\color{orangered}{ 21 }\\& & -20& 30& -10& \color{orangered}{100} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{1}&\color{blue}{-10}&\color{orangered}{121} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-3x^{2}+x-10 } $ with a remainder of $ \color{red}{ 121 } $.