Divide using synthetic division $$ ( x^{4}-8x^{3}+18x^{2}-24x+45 ) \div ( x-5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-8&18&-24&45\\& & 5& -15& 15& \color{black}{-45} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-8x^{3}+18x^{2}-24x+45 }{ x-5 } = \color{blue}{x^{3}-3x^{2}+3x-9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-8&18&-24&45\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-8&18&-24&45\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-8&18&-24&45\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 5 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -8 }&18&-24&45\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-8&18&-24&45\\& & 5& \color{blue}{-15} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}5&1&-8&\color{orangered}{ 18 }&-24&45\\& & 5& \color{orangered}{-15} & & \\ \hline &1&-3&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-8&18&-24&45\\& & 5& -15& \color{blue}{15} & \\ \hline &1&-3&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 15 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}5&1&-8&18&\color{orangered}{ -24 }&45\\& & 5& -15& \color{orangered}{15} & \\ \hline &1&-3&3&\color{orangered}{-9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-8&18&-24&45\\& & 5& -15& 15& \color{blue}{-45} \\ \hline &1&-3&3&\color{blue}{-9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&-8&18&-24&\color{orangered}{ 45 }\\& & 5& -15& 15& \color{orangered}{-45} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}+3x-9 } $ with a remainder of $ \color{red}{ 0 } $.