Divide using synthetic division $$ ( 6x^{4}+x^{3}-7x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&6&1&0&-7&3\\& & 18& 57& 171& \color{black}{492} \\ \hline &\color{blue}{6}&\color{blue}{19}&\color{blue}{57}&\color{blue}{164}&\color{orangered}{495} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+x^{3}-7x+3 }{ x-3 } = \color{blue}{6x^{3}+19x^{2}+57x+164} ~+~ \frac{ \color{red}{ 495 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&6&1&0&-7&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 6 }&1&0&-7&3\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&6&1&0&-7&3\\& & \color{blue}{18} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 18 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrr}3&6&\color{orangered}{ 1 }&0&-7&3\\& & \color{orangered}{18} & & & \\ \hline &6&\color{orangered}{19}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 19 } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&6&1&0&-7&3\\& & 18& \color{blue}{57} & & \\ \hline &6&\color{blue}{19}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 57 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrrr}3&6&1&\color{orangered}{ 0 }&-7&3\\& & 18& \color{orangered}{57} & & \\ \hline &6&19&\color{orangered}{57}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 57 } = \color{blue}{ 171 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&6&1&0&-7&3\\& & 18& 57& \color{blue}{171} & \\ \hline &6&19&\color{blue}{57}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 171 } = \color{orangered}{ 164 } $
$$ \begin{array}{c|rrrrr}3&6&1&0&\color{orangered}{ -7 }&3\\& & 18& 57& \color{orangered}{171} & \\ \hline &6&19&57&\color{orangered}{164}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 164 } = \color{blue}{ 492 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&6&1&0&-7&3\\& & 18& 57& 171& \color{blue}{492} \\ \hline &6&19&57&\color{blue}{164}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 492 } = \color{orangered}{ 495 } $
$$ \begin{array}{c|rrrrr}3&6&1&0&-7&\color{orangered}{ 3 }\\& & 18& 57& 171& \color{orangered}{492} \\ \hline &\color{blue}{6}&\color{blue}{19}&\color{blue}{57}&\color{blue}{164}&\color{orangered}{495} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+19x^{2}+57x+164 } $ with a remainder of $ \color{red}{ 495 } $.