Divide using synthetic division $$ ( 2x^{4}-15x^{3}+26x^{2}-60x+72 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&-15&26&-60&72\\& & 2& -13& 13& \color{black}{-47} \\ \hline &\color{blue}{2}&\color{blue}{-13}&\color{blue}{13}&\color{blue}{-47}&\color{orangered}{25} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-15x^{3}+26x^{2}-60x+72 }{ x-1 } = \color{blue}{2x^{3}-13x^{2}+13x-47} ~+~ \frac{ \color{red}{ 25 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-15&26&-60&72\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 2 }&-15&26&-60&72\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-15&26&-60&72\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 2 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ -15 }&26&-60&72\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{-13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-15&26&-60&72\\& & 2& \color{blue}{-13} & & \\ \hline &2&\color{blue}{-13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}1&2&-15&\color{orangered}{ 26 }&-60&72\\& & 2& \color{orangered}{-13} & & \\ \hline &2&-13&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 13 } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-15&26&-60&72\\& & 2& -13& \color{blue}{13} & \\ \hline &2&-13&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 13 } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrrr}1&2&-15&26&\color{orangered}{ -60 }&72\\& & 2& -13& \color{orangered}{13} & \\ \hline &2&-13&13&\color{orangered}{-47}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -47 \right) } = \color{blue}{ -47 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-15&26&-60&72\\& & 2& -13& 13& \color{blue}{-47} \\ \hline &2&-13&13&\color{blue}{-47}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 72 } + \color{orangered}{ \left( -47 \right) } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}1&2&-15&26&-60&\color{orangered}{ 72 }\\& & 2& -13& 13& \color{orangered}{-47} \\ \hline &\color{blue}{2}&\color{blue}{-13}&\color{blue}{13}&\color{blue}{-47}&\color{orangered}{25} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-13x^{2}+13x-47 } $ with a remainder of $ \color{red}{ 25 } $.