Divide using synthetic division $$ ( x^{4}-2x^{3}-80x+29 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-2&0&-80&29\\& & 5& 15& 75& \color{black}{-25} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{15}&\color{blue}{-5}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-80x+29 }{ x-5 } = \color{blue}{x^{3}+3x^{2}+15x-5} ~+~ \frac{ \color{red}{ 4 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&0&-80&29\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-2&0&-80&29\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&0&-80&29\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 5 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -2 }&0&-80&29\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&0&-80&29\\& & 5& \color{blue}{15} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}5&1&-2&\color{orangered}{ 0 }&-80&29\\& & 5& \color{orangered}{15} & & \\ \hline &1&3&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 15 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&0&-80&29\\& & 5& 15& \color{blue}{75} & \\ \hline &1&3&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -80 } + \color{orangered}{ 75 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}5&1&-2&0&\color{orangered}{ -80 }&29\\& & 5& 15& \color{orangered}{75} & \\ \hline &1&3&15&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&0&-80&29\\& & 5& 15& 75& \color{blue}{-25} \\ \hline &1&3&15&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&1&-2&0&-80&\color{orangered}{ 29 }\\& & 5& 15& 75& \color{orangered}{-25} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{15}&\color{blue}{-5}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}+15x-5 } $ with a remainder of $ \color{red}{ 4 } $.