Divide using synthetic division $$ ( 2x^{5}-102x^{3}-38x^{2}+1020x-450 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&2&0&-102&-38&1020&-450\\& & -10& 50& 260& -1110& \color{black}{450} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-52}&\color{blue}{222}&\color{blue}{-90}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-102x^{3}-38x^{2}+1020x-450 }{ x+5 } = \color{blue}{2x^{4}-10x^{3}-52x^{2}+222x-90} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 2 }&0&-102&-38&1020&-450\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & \color{blue}{-10} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrrr}-5&2&\color{orangered}{ 0 }&-102&-38&1020&-450\\& & \color{orangered}{-10} & & & & \\ \hline &2&\color{orangered}{-10}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & -10& \color{blue}{50} & & & \\ \hline &2&\color{blue}{-10}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -102 } + \color{orangered}{ 50 } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&\color{orangered}{ -102 }&-38&1020&-450\\& & -10& \color{orangered}{50} & & & \\ \hline &2&-10&\color{orangered}{-52}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -52 \right) } = \color{blue}{ 260 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & -10& 50& \color{blue}{260} & & \\ \hline &2&-10&\color{blue}{-52}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ 260 } = \color{orangered}{ 222 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-102&\color{orangered}{ -38 }&1020&-450\\& & -10& 50& \color{orangered}{260} & & \\ \hline &2&-10&-52&\color{orangered}{222}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 222 } = \color{blue}{ -1110 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & -10& 50& 260& \color{blue}{-1110} & \\ \hline &2&-10&-52&\color{blue}{222}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1020 } + \color{orangered}{ \left( -1110 \right) } = \color{orangered}{ -90 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-102&-38&\color{orangered}{ 1020 }&-450\\& & -10& 50& 260& \color{orangered}{-1110} & \\ \hline &2&-10&-52&222&\color{orangered}{-90}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -90 \right) } = \color{blue}{ 450 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&2&0&-102&-38&1020&-450\\& & -10& 50& 260& -1110& \color{blue}{450} \\ \hline &2&-10&-52&222&\color{blue}{-90}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -450 } + \color{orangered}{ 450 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-5&2&0&-102&-38&1020&\color{orangered}{ -450 }\\& & -10& 50& 260& -1110& \color{orangered}{450} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{-52}&\color{blue}{222}&\color{blue}{-90}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-10x^{3}-52x^{2}+222x-90 } $ with a remainder of $ \color{red}{ 0 } $.