Divide using synthetic division $$ ( x^{4}+x^{3}-7x^{2}-2x+8 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&1&-7&-2&8\\& & 5& 30& 115& \color{black}{565} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{23}&\color{blue}{113}&\color{orangered}{573} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}-7x^{2}-2x+8 }{ x-5 } = \color{blue}{x^{3}+6x^{2}+23x+113} ~+~ \frac{ \color{red}{ 573 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&-7&-2&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&1&-7&-2&8\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&-7&-2&8\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 5 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 1 }&-7&-2&8\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&-7&-2&8\\& & 5& \color{blue}{30} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 30 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrrr}5&1&1&\color{orangered}{ -7 }&-2&8\\& & 5& \color{orangered}{30} & & \\ \hline &1&6&\color{orangered}{23}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 23 } = \color{blue}{ 115 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&-7&-2&8\\& & 5& 30& \color{blue}{115} & \\ \hline &1&6&\color{blue}{23}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 115 } = \color{orangered}{ 113 } $
$$ \begin{array}{c|rrrrr}5&1&1&-7&\color{orangered}{ -2 }&8\\& & 5& 30& \color{orangered}{115} & \\ \hline &1&6&23&\color{orangered}{113}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 113 } = \color{blue}{ 565 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&1&-7&-2&8\\& & 5& 30& 115& \color{blue}{565} \\ \hline &1&6&23&\color{blue}{113}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 565 } = \color{orangered}{ 573 } $
$$ \begin{array}{c|rrrrr}5&1&1&-7&-2&\color{orangered}{ 8 }\\& & 5& 30& 115& \color{orangered}{565} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{23}&\color{blue}{113}&\color{orangered}{573} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+6x^{2}+23x+113 } $ with a remainder of $ \color{red}{ 573 } $.