Divide using synthetic division $$ ( 4x^{4}+6x^{3}-15x^{2}+1 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&6&-15&0&1\\& & 4& 10& -5& \color{black}{-5} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{-5}&\color{blue}{-5}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+6x^{3}-15x^{2}+1 }{ x-1 } = \color{blue}{4x^{3}+10x^{2}-5x-5} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&6&-15&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&6&-15&0&1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&6&-15&0&1\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 4 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ 6 }&-15&0&1\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&6&-15&0&1\\& & 4& \color{blue}{10} & & \\ \hline &4&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 10 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&4&6&\color{orangered}{ -15 }&0&1\\& & 4& \color{orangered}{10} & & \\ \hline &4&10&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&6&-15&0&1\\& & 4& 10& \color{blue}{-5} & \\ \hline &4&10&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&4&6&-15&\color{orangered}{ 0 }&1\\& & 4& 10& \color{orangered}{-5} & \\ \hline &4&10&-5&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&6&-15&0&1\\& & 4& 10& -5& \color{blue}{-5} \\ \hline &4&10&-5&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}1&4&6&-15&0&\color{orangered}{ 1 }\\& & 4& 10& -5& \color{orangered}{-5} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{-5}&\color{blue}{-5}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+10x^{2}-5x-5 } $ with a remainder of $ \color{red}{ -4 } $.