Divide using synthetic division $$ ( x^{4}+x^{3}-6x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&1&0&-6&3\\& & 3& 12& 36& \color{black}{90} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{12}&\color{blue}{30}&\color{orangered}{93} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}-6x+3 }{ x-3 } = \color{blue}{x^{3}+4x^{2}+12x+30} ~+~ \frac{ \color{red}{ 93 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&1&0&-6&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&1&0&-6&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&1&0&-6&3\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 3 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 1 }&0&-6&3\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&1&0&-6&3\\& & 3& \color{blue}{12} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&1&1&\color{orangered}{ 0 }&-6&3\\& & 3& \color{orangered}{12} & & \\ \hline &1&4&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&1&0&-6&3\\& & 3& 12& \color{blue}{36} & \\ \hline &1&4&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 36 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}3&1&1&0&\color{orangered}{ -6 }&3\\& & 3& 12& \color{orangered}{36} & \\ \hline &1&4&12&\color{orangered}{30}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 30 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&1&0&-6&3\\& & 3& 12& 36& \color{blue}{90} \\ \hline &1&4&12&\color{blue}{30}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 90 } = \color{orangered}{ 93 } $
$$ \begin{array}{c|rrrrr}3&1&1&0&-6&\color{orangered}{ 3 }\\& & 3& 12& 36& \color{orangered}{90} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{12}&\color{blue}{30}&\color{orangered}{93} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}+12x+30 } $ with a remainder of $ \color{red}{ 93 } $.