Divide using synthetic division $$ ( 4x^{2}+x-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&4&1&-3\\& & 12& \color{black}{39} \\ \hline &\color{blue}{4}&\color{blue}{13}&\color{orangered}{36} \end{array} $$The solution is:
$$ \frac{ 4x^{2}+x-3 }{ x-3 } = \color{blue}{4x+13} ~+~ \frac{ \color{red}{ 36 } }{ x-3 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&1&-3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 4 }&1&-3\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&1&-3\\& & \color{blue}{12} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 12 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrr}3&4&\color{orangered}{ 1 }&-3\\& & \color{orangered}{12} & \\ \hline &4&\color{orangered}{13}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&1&-3\\& & 12& \color{blue}{39} \\ \hline &4&\color{blue}{13}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 39 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrr}3&4&1&\color{orangered}{ -3 }\\& & 12& \color{orangered}{39} \\ \hline &\color{blue}{4}&\color{blue}{13}&\color{orangered}{36} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x+13 } $ with a remainder of $ \color{red}{ 36 } $.