Divide using synthetic division $$ ( 7x^{4}+6x^{3}-36x^{2}-30x+5 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&7&6&-36&-30&5\\& & 7& 13& -23& \color{black}{-53} \\ \hline &\color{blue}{7}&\color{blue}{13}&\color{blue}{-23}&\color{blue}{-53}&\color{orangered}{-48} \end{array} $$The solution is:
$$ \frac{ 7x^{4}+6x^{3}-36x^{2}-30x+5 }{ x-1 } = \color{blue}{7x^{3}+13x^{2}-23x-53} \color{red}{~-~} \frac{ \color{red}{ 48 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&6&-36&-30&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 7 }&6&-36&-30&5\\& & & & & \\ \hline &\color{orangered}{7}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&6&-36&-30&5\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{7}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 7 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}1&7&\color{orangered}{ 6 }&-36&-30&5\\& & \color{orangered}{7} & & & \\ \hline &7&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 13 } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&6&-36&-30&5\\& & 7& \color{blue}{13} & & \\ \hline &7&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 13 } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrr}1&7&6&\color{orangered}{ -36 }&-30&5\\& & 7& \color{orangered}{13} & & \\ \hline &7&13&\color{orangered}{-23}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ -23 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&6&-36&-30&5\\& & 7& 13& \color{blue}{-23} & \\ \hline &7&13&\color{blue}{-23}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ \left( -23 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrr}1&7&6&-36&\color{orangered}{ -30 }&5\\& & 7& 13& \color{orangered}{-23} & \\ \hline &7&13&-23&\color{orangered}{-53}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -53 \right) } = \color{blue}{ -53 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&7&6&-36&-30&5\\& & 7& 13& -23& \color{blue}{-53} \\ \hline &7&13&-23&\color{blue}{-53}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -53 \right) } = \color{orangered}{ -48 } $
$$ \begin{array}{c|rrrrr}1&7&6&-36&-30&\color{orangered}{ 5 }\\& & 7& 13& -23& \color{orangered}{-53} \\ \hline &\color{blue}{7}&\color{blue}{13}&\color{blue}{-23}&\color{blue}{-53}&\color{orangered}{-48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{3}+13x^{2}-23x-53 } $ with a remainder of $ \color{red}{ -48 } $.