Divide using synthetic division $$ ( 4x^{3}+5x^{2}+8 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&5&0&8\\& & 12& 51& \color{black}{153} \\ \hline &\color{blue}{4}&\color{blue}{17}&\color{blue}{51}&\color{orangered}{161} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+5x^{2}+8 }{ x-3 } = \color{blue}{4x^{2}+17x+51} ~+~ \frac{ \color{red}{ 161 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&5&0&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&5&0&8\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&5&0&8\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 12 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ 5 }&0&8\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 17 } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&5&0&8\\& & 12& \color{blue}{51} & \\ \hline &4&\color{blue}{17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 51 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrr}3&4&5&\color{orangered}{ 0 }&8\\& & 12& \color{orangered}{51} & \\ \hline &4&17&\color{orangered}{51}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 51 } = \color{blue}{ 153 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&5&0&8\\& & 12& 51& \color{blue}{153} \\ \hline &4&17&\color{blue}{51}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 153 } = \color{orangered}{ 161 } $
$$ \begin{array}{c|rrrr}3&4&5&0&\color{orangered}{ 8 }\\& & 12& 51& \color{orangered}{153} \\ \hline &\color{blue}{4}&\color{blue}{17}&\color{blue}{51}&\color{orangered}{161} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+17x+51 } $ with a remainder of $ \color{red}{ 161 } $.