Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&7&-5&-21&15\\& & 7& 2& \color{black}{-19} \\ \hline &\color{blue}{7}&\color{blue}{2}&\color{blue}{-19}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x-1 } = \color{blue}{7x^{2}+2x-19} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-5&-21&15\\& & \color{blue}{7} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 7 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}1&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{7} & & \\ \hline &7&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-5&-21&15\\& & 7& \color{blue}{2} & \\ \hline &7&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 2 } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}1&7&-5&\color{orangered}{ -21 }&15\\& & 7& \color{orangered}{2} & \\ \hline &7&2&\color{orangered}{-19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ -19 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-5&-21&15\\& & 7& 2& \color{blue}{-19} \\ \hline &7&2&\color{blue}{-19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -19 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}1&7&-5&-21&\color{orangered}{ 15 }\\& & 7& 2& \color{orangered}{-19} \\ \hline &\color{blue}{7}&\color{blue}{2}&\color{blue}{-19}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+2x-19 } $ with a remainder of $ \color{red}{ -4 } $.