Divide using synthetic division $$ ( 4x^{5}-10x^{4}+8x^{3}+x^{2}-8 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&4&-10&8&1&0&-8\\& & 4& -6& 2& 3& \color{black}{3} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{3}&\color{blue}{3}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ 4x^{5}-10x^{4}+8x^{3}+x^{2}-8 }{ x-1 } = \color{blue}{4x^{4}-6x^{3}+2x^{2}+3x+3} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 4 }&-10&8&1&0&-8\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & \color{blue}{4} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 4 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}1&4&\color{orangered}{ -10 }&8&1&0&-8\\& & \color{orangered}{4} & & & & \\ \hline &4&\color{orangered}{-6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & 4& \color{blue}{-6} & & & \\ \hline &4&\color{blue}{-6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}1&4&-10&\color{orangered}{ 8 }&1&0&-8\\& & 4& \color{orangered}{-6} & & & \\ \hline &4&-6&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & 4& -6& \color{blue}{2} & & \\ \hline &4&-6&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}1&4&-10&8&\color{orangered}{ 1 }&0&-8\\& & 4& -6& \color{orangered}{2} & & \\ \hline &4&-6&2&\color{orangered}{3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & 4& -6& 2& \color{blue}{3} & \\ \hline &4&-6&2&\color{blue}{3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}1&4&-10&8&1&\color{orangered}{ 0 }&-8\\& & 4& -6& 2& \color{orangered}{3} & \\ \hline &4&-6&2&3&\color{orangered}{3}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&4&-10&8&1&0&-8\\& & 4& -6& 2& 3& \color{blue}{3} \\ \hline &4&-6&2&3&\color{blue}{3}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 3 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}1&4&-10&8&1&0&\color{orangered}{ -8 }\\& & 4& -6& 2& 3& \color{orangered}{3} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{3}&\color{blue}{3}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}-6x^{3}+2x^{2}+3x+3 } $ with a remainder of $ \color{red}{ -5 } $.