Divide using synthetic division $$ ( 7x^{3}+x^{2}-5x-6 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&7&1&-5&-6\\& & 14& 30& \color{black}{50} \\ \hline &\color{blue}{7}&\color{blue}{15}&\color{blue}{25}&\color{orangered}{44} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+x^{2}-5x-6 }{ x-2 } = \color{blue}{7x^{2}+15x+25} ~+~ \frac{ \color{red}{ 44 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&1&-5&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 7 }&1&-5&-6\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&1&-5&-6\\& & \color{blue}{14} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 14 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}2&7&\color{orangered}{ 1 }&-5&-6\\& & \color{orangered}{14} & & \\ \hline &7&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&1&-5&-6\\& & 14& \color{blue}{30} & \\ \hline &7&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 30 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}2&7&1&\color{orangered}{ -5 }&-6\\& & 14& \color{orangered}{30} & \\ \hline &7&15&\color{orangered}{25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 25 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&1&-5&-6\\& & 14& 30& \color{blue}{50} \\ \hline &7&15&\color{blue}{25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 50 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrr}2&7&1&-5&\color{orangered}{ -6 }\\& & 14& 30& \color{orangered}{50} \\ \hline &\color{blue}{7}&\color{blue}{15}&\color{blue}{25}&\color{orangered}{44} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+15x+25 } $ with a remainder of $ \color{red}{ 44 } $.