Divide using synthetic division $$ ( x^{3}+4x^{2}-5x+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&4&-5&1\\& & 3& 21& \color{black}{48} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{16}&\color{orangered}{49} \end{array} $$The solution is:
$$ \frac{ x^{3}+4x^{2}-5x+1 }{ x-3 } = \color{blue}{x^{2}+7x+16} ~+~ \frac{ \color{red}{ 49 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&-5&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&4&-5&1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&-5&1\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 3 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 4 }&-5&1\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&-5&1\\& & 3& \color{blue}{21} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 21 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}3&1&4&\color{orangered}{ -5 }&1\\& & 3& \color{orangered}{21} & \\ \hline &1&7&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&-5&1\\& & 3& 21& \color{blue}{48} \\ \hline &1&7&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 48 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}3&1&4&-5&\color{orangered}{ 1 }\\& & 3& 21& \color{orangered}{48} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{16}&\color{orangered}{49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+7x+16 } $ with a remainder of $ \color{red}{ 49 } $.