Divide using synthetic division $$ ( x^{3}+2x^{2}-3x+5 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&2&-3&5\\& & 3& 15& \color{black}{36} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{12}&\color{orangered}{41} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-3x+5 }{ x-3 } = \color{blue}{x^{2}+5x+12} ~+~ \frac{ \color{red}{ 41 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&2&-3&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&2&-3&5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&2&-3&5\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 2 }&-3&5\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&2&-3&5\\& & 3& \color{blue}{15} & \\ \hline &1&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 15 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}3&1&2&\color{orangered}{ -3 }&5\\& & 3& \color{orangered}{15} & \\ \hline &1&5&\color{orangered}{12}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&2&-3&5\\& & 3& 15& \color{blue}{36} \\ \hline &1&5&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 36 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrr}3&1&2&-3&\color{orangered}{ 5 }\\& & 3& 15& \color{orangered}{36} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{12}&\color{orangered}{41} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+5x+12 } $ with a remainder of $ \color{red}{ 41 } $.