Divide using synthetic division $$ ( 4x^{4}+2x^{3}-x^{2}+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&2&-1&0&3\\& & 12& 42& 123& \color{black}{369} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{blue}{41}&\color{blue}{123}&\color{orangered}{372} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+2x^{3}-x^{2}+3 }{ x-3 } = \color{blue}{4x^{3}+14x^{2}+41x+123} ~+~ \frac{ \color{red}{ 372 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&2&-1&0&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&2&-1&0&3\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&2&-1&0&3\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 12 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ 2 }&-1&0&3\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{14}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 14 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&2&-1&0&3\\& & 12& \color{blue}{42} & & \\ \hline &4&\color{blue}{14}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 42 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrrr}3&4&2&\color{orangered}{ -1 }&0&3\\& & 12& \color{orangered}{42} & & \\ \hline &4&14&\color{orangered}{41}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 41 } = \color{blue}{ 123 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&2&-1&0&3\\& & 12& 42& \color{blue}{123} & \\ \hline &4&14&\color{blue}{41}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 123 } = \color{orangered}{ 123 } $
$$ \begin{array}{c|rrrrr}3&4&2&-1&\color{orangered}{ 0 }&3\\& & 12& 42& \color{orangered}{123} & \\ \hline &4&14&41&\color{orangered}{123}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 123 } = \color{blue}{ 369 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&2&-1&0&3\\& & 12& 42& 123& \color{blue}{369} \\ \hline &4&14&41&\color{blue}{123}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 369 } = \color{orangered}{ 372 } $
$$ \begin{array}{c|rrrrr}3&4&2&-1&0&\color{orangered}{ 3 }\\& & 12& 42& 123& \color{orangered}{369} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{blue}{41}&\color{blue}{123}&\color{orangered}{372} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+14x^{2}+41x+123 } $ with a remainder of $ \color{red}{ 372 } $.