Divide using synthetic division $$ ( 4x^{4}-x^{3}+20x-5 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&-1&0&20&-5\\& & 4& 3& 3& \color{black}{23} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{3}&\color{blue}{23}&\color{orangered}{18} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-x^{3}+20x-5 }{ x-1 } = \color{blue}{4x^{3}+3x^{2}+3x+23} ~+~ \frac{ \color{red}{ 18 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&0&20&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&-1&0&20&-5\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&0&20&-5\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 4 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ -1 }&0&20&-5\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&0&20&-5\\& & 4& \color{blue}{3} & & \\ \hline &4&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&4&-1&\color{orangered}{ 0 }&20&-5\\& & 4& \color{orangered}{3} & & \\ \hline &4&3&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&0&20&-5\\& & 4& 3& \color{blue}{3} & \\ \hline &4&3&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 3 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrrr}1&4&-1&0&\color{orangered}{ 20 }&-5\\& & 4& 3& \color{orangered}{3} & \\ \hline &4&3&3&\color{orangered}{23}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 23 } = \color{blue}{ 23 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&-1&0&20&-5\\& & 4& 3& 3& \color{blue}{23} \\ \hline &4&3&3&\color{blue}{23}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 23 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}1&4&-1&0&20&\color{orangered}{ -5 }\\& & 4& 3& 3& \color{orangered}{23} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{3}&\color{blue}{23}&\color{orangered}{18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+3x^{2}+3x+23 } $ with a remainder of $ \color{red}{ 18 } $.