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