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