Divide using synthetic division $$ ( 4x^{3}+19x^{2}-61x+14 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&4&19&-61&14\\& & 28& 329& \color{black}{1876} \\ \hline &\color{blue}{4}&\color{blue}{47}&\color{blue}{268}&\color{orangered}{1890} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+19x^{2}-61x+14 }{ x-7 } = \color{blue}{4x^{2}+47x+268} ~+~ \frac{ \color{red}{ 1890 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&4&19&-61&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 4 }&19&-61&14\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 4 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&4&19&-61&14\\& & \color{blue}{28} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 28 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}7&4&\color{orangered}{ 19 }&-61&14\\& & \color{orangered}{28} & & \\ \hline &4&\color{orangered}{47}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 47 } = \color{blue}{ 329 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&4&19&-61&14\\& & 28& \color{blue}{329} & \\ \hline &4&\color{blue}{47}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -61 } + \color{orangered}{ 329 } = \color{orangered}{ 268 } $
$$ \begin{array}{c|rrrr}7&4&19&\color{orangered}{ -61 }&14\\& & 28& \color{orangered}{329} & \\ \hline &4&47&\color{orangered}{268}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 268 } = \color{blue}{ 1876 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&4&19&-61&14\\& & 28& 329& \color{blue}{1876} \\ \hline &4&47&\color{blue}{268}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 1876 } = \color{orangered}{ 1890 } $
$$ \begin{array}{c|rrrr}7&4&19&-61&\color{orangered}{ 14 }\\& & 28& 329& \color{orangered}{1876} \\ \hline &\color{blue}{4}&\color{blue}{47}&\color{blue}{268}&\color{orangered}{1890} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+47x+268 } $ with a remainder of $ \color{red}{ 1890 } $.