Divide using synthetic division $$ ( x^{4}-8x^{3}+18x^{2}-24x+45 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-8&18&-24&45\\& & 2& -12& 12& \color{black}{-24} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{6}&\color{blue}{-12}&\color{orangered}{21} \end{array} $$The solution is:
$$ \frac{ x^{4}-8x^{3}+18x^{2}-24x+45 }{ x-2 } = \color{blue}{x^{3}-6x^{2}+6x-12} ~+~ \frac{ \color{red}{ 21 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-8&18&-24&45\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-8&18&-24&45\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-8&18&-24&45\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 2 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -8 }&18&-24&45\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-8&18&-24&45\\& & 2& \color{blue}{-12} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&1&-8&\color{orangered}{ 18 }&-24&45\\& & 2& \color{orangered}{-12} & & \\ \hline &1&-6&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-8&18&-24&45\\& & 2& -12& \color{blue}{12} & \\ \hline &1&-6&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 12 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}2&1&-8&18&\color{orangered}{ -24 }&45\\& & 2& -12& \color{orangered}{12} & \\ \hline &1&-6&6&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-8&18&-24&45\\& & 2& -12& 12& \color{blue}{-24} \\ \hline &1&-6&6&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}2&1&-8&18&-24&\color{orangered}{ 45 }\\& & 2& -12& 12& \color{orangered}{-24} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{6}&\color{blue}{-12}&\color{orangered}{21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+6x-12 } $ with a remainder of $ \color{red}{ 21 } $.