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