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