Divide using synthetic division $$ ( x^{3}-17x^{2}+45x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-17&45&0\\& & 3& -42& \color{black}{9} \\ \hline &\color{blue}{1}&\color{blue}{-14}&\color{blue}{3}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ x^{3}-17x^{2}+45x }{ x-3 } = \color{blue}{x^{2}-14x+3} ~+~ \frac{ \color{red}{ 9 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-17&45&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-17&45&0\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-17&45&0\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 3 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -17 }&45&0\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-17&45&0\\& & 3& \color{blue}{-42} & \\ \hline &1&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&1&-17&\color{orangered}{ 45 }&0\\& & 3& \color{orangered}{-42} & \\ \hline &1&-14&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-17&45&0\\& & 3& -42& \color{blue}{9} \\ \hline &1&-14&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}3&1&-17&45&\color{orangered}{ 0 }\\& & 3& -42& \color{orangered}{9} \\ \hline &\color{blue}{1}&\color{blue}{-14}&\color{blue}{3}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-14x+3 } $ with a remainder of $ \color{red}{ 9 } $.