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