Divide using synthetic division $$ ( 4x^{6}-3x^{4}+x^{2}+5 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}1&4&0&-3&0&1&0&5\\& & 4& 4& 1& 1& 2& \color{black}{2} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{1}&\color{blue}{1}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 4x^{6}-3x^{4}+x^{2}+5 }{ x-1 } = \color{blue}{4x^{5}+4x^{4}+x^{3}+x^{2}+2x+2} ~+~ \frac{ \color{red}{ 7 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}1&\color{orangered}{ 4 }&0&-3&0&1&0&5\\& & & & & & & \\ \hline &\color{orangered}{4}&&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & \color{blue}{4} & & & & & \\ \hline &\color{blue}{4}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrrr}1&4&\color{orangered}{ 0 }&-3&0&1&0&5\\& & \color{orangered}{4} & & & & & \\ \hline &4&\color{orangered}{4}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & 4& \color{blue}{4} & & & & \\ \hline &4&\color{blue}{4}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 4 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrrr}1&4&0&\color{orangered}{ -3 }&0&1&0&5\\& & 4& \color{orangered}{4} & & & & \\ \hline &4&4&\color{orangered}{1}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & 4& 4& \color{blue}{1} & & & \\ \hline &4&4&\color{blue}{1}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrrr}1&4&0&-3&\color{orangered}{ 0 }&1&0&5\\& & 4& 4& \color{orangered}{1} & & & \\ \hline &4&4&1&\color{orangered}{1}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & 4& 4& 1& \color{blue}{1} & & \\ \hline &4&4&1&\color{blue}{1}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 1 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrrr}1&4&0&-3&0&\color{orangered}{ 1 }&0&5\\& & 4& 4& 1& \color{orangered}{1} & & \\ \hline &4&4&1&1&\color{orangered}{2}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & 4& 4& 1& 1& \color{blue}{2} & \\ \hline &4&4&1&1&\color{blue}{2}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrrr}1&4&0&-3&0&1&\color{orangered}{ 0 }&5\\& & 4& 4& 1& 1& \color{orangered}{2} & \\ \hline &4&4&1&1&2&\color{orangered}{2}& \end{array} $$Step 12 : 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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{1}&4&0&-3&0&1&0&5\\& & 4& 4& 1& 1& 2& \color{blue}{2} \\ \hline &4&4&1&1&2&\color{blue}{2}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 2 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrrr}1&4&0&-3&0&1&0&\color{orangered}{ 5 }\\& & 4& 4& 1& 1& 2& \color{orangered}{2} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{1}&\color{blue}{1}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{5}+4x^{4}+x^{3}+x^{2}+2x+2 } $ with a remainder of $ \color{red}{ 7 } $.