Divide using synthetic division $$ ( x^{3}-11x^{2}+39x-45 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-11&39&-45\\& & 3& -24& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{15}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-11x^{2}+39x-45 }{ x-3 } = \color{blue}{x^{2}-8x+15} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-11&39&-45\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-11&39&-45\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-11&39&-45\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 3 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -11 }&39&-45\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-11&39&-45\\& & 3& \color{blue}{-24} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 39 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}3&1&-11&\color{orangered}{ 39 }&-45\\& & 3& \color{orangered}{-24} & \\ \hline &1&-8&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-11&39&-45\\& & 3& -24& \color{blue}{45} \\ \hline &1&-8&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 45 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&-11&39&\color{orangered}{ -45 }\\& & 3& -24& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{15}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-8x+15 } $ with a remainder of $ \color{red}{ 0 } $.