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