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