Divide using synthetic division $$ ( 7x^{3}+18x^{2}+30x-109 ) \div ( 7x-10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}10&7&18&30&-109\\& & 70& 880& \color{black}{9100} \\ \hline &\color{blue}{7}&\color{blue}{88}&\color{blue}{910}&\color{orangered}{8991} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+18x^{2}+30x-109 }{ x-10 } = \color{blue}{7x^{2}+88x+910} ~+~ \frac{ \color{red}{ 8991 } }{ x-10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -10 = 0 $ ( $ x = \color{blue}{ 10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{10}&7&18&30&-109\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}10&\color{orangered}{ 7 }&18&30&-109\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 7 } = \color{blue}{ 70 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&7&18&30&-109\\& & \color{blue}{70} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ 70 } = \color{orangered}{ 88 } $
$$ \begin{array}{c|rrrr}10&7&\color{orangered}{ 18 }&30&-109\\& & \color{orangered}{70} & & \\ \hline &7&\color{orangered}{88}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 88 } = \color{blue}{ 880 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&7&18&30&-109\\& & 70& \color{blue}{880} & \\ \hline &7&\color{blue}{88}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ 880 } = \color{orangered}{ 910 } $
$$ \begin{array}{c|rrrr}10&7&18&\color{orangered}{ 30 }&-109\\& & 70& \color{orangered}{880} & \\ \hline &7&88&\color{orangered}{910}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 910 } = \color{blue}{ 9100 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&7&18&30&-109\\& & 70& 880& \color{blue}{9100} \\ \hline &7&88&\color{blue}{910}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -109 } + \color{orangered}{ 9100 } = \color{orangered}{ 8991 } $
$$ \begin{array}{c|rrrr}10&7&18&30&\color{orangered}{ -109 }\\& & 70& 880& \color{orangered}{9100} \\ \hline &\color{blue}{7}&\color{blue}{88}&\color{blue}{910}&\color{orangered}{8991} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+88x+910 } $ with a remainder of $ \color{red}{ 8991 } $.