Divide using synthetic division $$ ( x^{3}-18x^{2}+99x-152 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-18&99&-152\\& & 3& -45& \color{black}{162} \\ \hline &\color{blue}{1}&\color{blue}{-15}&\color{blue}{54}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{3}-18x^{2}+99x-152 }{ x-3 } = \color{blue}{x^{2}-15x+54} ~+~ \frac{ \color{red}{ 10 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-18&99&-152\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-18&99&-152\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-18&99&-152\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 3 } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -18 }&99&-152\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-18&99&-152\\& & 3& \color{blue}{-45} & \\ \hline &1&\color{blue}{-15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 99 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrr}3&1&-18&\color{orangered}{ 99 }&-152\\& & 3& \color{orangered}{-45} & \\ \hline &1&-15&\color{orangered}{54}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 54 } = \color{blue}{ 162 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-18&99&-152\\& & 3& -45& \color{blue}{162} \\ \hline &1&-15&\color{blue}{54}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -152 } + \color{orangered}{ 162 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}3&1&-18&99&\color{orangered}{ -152 }\\& & 3& -45& \color{orangered}{162} \\ \hline &\color{blue}{1}&\color{blue}{-15}&\color{blue}{54}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-15x+54 } $ with a remainder of $ \color{red}{ 10 } $.