Divide using synthetic division $$ ( x^{5}-18x^{3}+39x-18 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}4&1&0&-18&0&39&-18\\& & 4& 16& -8& -32& \color{black}{28} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{7}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{5}-18x^{3}+39x-18 }{ x-4 } = \color{blue}{x^{4}+4x^{3}-2x^{2}-8x+7} ~+~ \frac{ \color{red}{ 10 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}4&\color{orangered}{ 1 }&0&-18&0&39&-18\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & \color{blue}{4} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}4&1&\color{orangered}{ 0 }&-18&0&39&-18\\& & \color{orangered}{4} & & & & \\ \hline &1&\color{orangered}{4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & 4& \color{blue}{16} & & & \\ \hline &1&\color{blue}{4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 16 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}4&1&0&\color{orangered}{ -18 }&0&39&-18\\& & 4& \color{orangered}{16} & & & \\ \hline &1&4&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & 4& 16& \color{blue}{-8} & & \\ \hline &1&4&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}4&1&0&-18&\color{orangered}{ 0 }&39&-18\\& & 4& 16& \color{orangered}{-8} & & \\ \hline &1&4&-2&\color{orangered}{-8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & 4& 16& -8& \color{blue}{-32} & \\ \hline &1&4&-2&\color{blue}{-8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 39 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}4&1&0&-18&0&\color{orangered}{ 39 }&-18\\& & 4& 16& -8& \color{orangered}{-32} & \\ \hline &1&4&-2&-8&\color{orangered}{7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 7 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{4}&1&0&-18&0&39&-18\\& & 4& 16& -8& -32& \color{blue}{28} \\ \hline &1&4&-2&-8&\color{blue}{7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 28 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}4&1&0&-18&0&39&\color{orangered}{ -18 }\\& & 4& 16& -8& -32& \color{orangered}{28} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-2}&\color{blue}{-8}&\color{blue}{7}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+4x^{3}-2x^{2}-8x+7 } $ with a remainder of $ \color{red}{ 10 } $.