Divide using synthetic division $$ ( 2x^{5}-103x^{3}-38x^{2}+1020x-450 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&2&0&-103&-38&1020&-450\\& & -10& 50& 265& -1135& \color{black}{575} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-53}&\color{blue}{227}&\color{blue}{-115}&\color{orangered}{125} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-103x^{3}-38x^{2}+1020x-450 }{ x+5 } = \color{blue}{2x^{4}-10x^{3}-53x^{2}+227x-115} ~+~ \frac{ \color{red}{ 125 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 2 }&0&-103&-38&1020&-450\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & \color{blue}{-10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrrr}-5&2&\color{orangered}{ 0 }&-103&-38&1020&-450\\& & \color{orangered}{-10} & & & & \\ \hline &2&\color{orangered}{-10}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & -10& \color{blue}{50} & & & \\ \hline &2&\color{blue}{-10}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -103 } + \color{orangered}{ 50 } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&\color{orangered}{ -103 }&-38&1020&-450\\& & -10& \color{orangered}{50} & & & \\ \hline &2&-10&\color{orangered}{-53}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -53 \right) } = \color{blue}{ 265 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & -10& 50& \color{blue}{265} & & \\ \hline &2&-10&\color{blue}{-53}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ 265 } = \color{orangered}{ 227 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-103&\color{orangered}{ -38 }&1020&-450\\& & -10& 50& \color{orangered}{265} & & \\ \hline &2&-10&-53&\color{orangered}{227}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 227 } = \color{blue}{ -1135 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & -10& 50& 265& \color{blue}{-1135} & \\ \hline &2&-10&-53&\color{blue}{227}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1020 } + \color{orangered}{ \left( -1135 \right) } = \color{orangered}{ -115 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-103&-38&\color{orangered}{ 1020 }&-450\\& & -10& 50& 265& \color{orangered}{-1135} & \\ \hline &2&-10&-53&227&\color{orangered}{-115}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -115 \right) } = \color{blue}{ 575 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-103&-38&1020&-450\\& & -10& 50& 265& -1135& \color{blue}{575} \\ \hline &2&-10&-53&227&\color{blue}{-115}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -450 } + \color{orangered}{ 575 } = \color{orangered}{ 125 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-103&-38&1020&\color{orangered}{ -450 }\\& & -10& 50& 265& -1135& \color{orangered}{575} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-53}&\color{blue}{227}&\color{blue}{-115}&\color{orangered}{125} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-10x^{3}-53x^{2}+227x-115 } $ with a remainder of $ \color{red}{ 125 } $.