Divide using synthetic division $$ ( 4x^{4}-7x^{3}+7x^{2}-12x+15 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&-7&7&-12&15\\& & 4& -3& 4& \color{black}{-8} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-8}&\color{orangered}{7} \end{array} $$The solution is:
$$ \dfrac{ 4x^{4}-7x^{3}+7x^{2}-12x+15 }{ x-1 } = \color{blue}{4x^{3}-3x^{2}+4x-8} ~+~ \dfrac{ \color{red}{ 7 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-7&7&-12&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&-7&7&-12&15\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-7&7&-12&15\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ -7 }&7&-12&15\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-7&7&-12&15\\& & 4& \color{blue}{-3} & & \\ \hline &4&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}1&4&-7&\color{orangered}{ 7 }&-12&15\\& & 4& \color{orangered}{-3} & & \\ \hline &4&-3&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-7&7&-12&15\\& & 4& -3& \color{blue}{4} & \\ \hline &4&-3&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 4 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}1&4&-7&7&\color{orangered}{ -12 }&15\\& & 4& -3& \color{orangered}{4} & \\ \hline &4&-3&4&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-7&7&-12&15\\& & 4& -3& 4& \color{blue}{-8} \\ \hline &4&-3&4&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}1&4&-7&7&-12&\color{orangered}{ 15 }\\& & 4& -3& 4& \color{orangered}{-8} \\ \hline &\color{blue}{4}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-8}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-3x^{2}+4x-8 } $ with a remainder of $ \color{red}{ 7 } $.