Divide using synthetic division $$ ( 5x^{4}-22x^{3}-8x^{2}+77x-52 ) \div ( 4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&5&-22&-8&77&-52\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{-22}&\color{blue}{-8}&\color{blue}{77}&\color{orangered}{-52} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-22x^{3}-8x^{2}+77x-52 }{ x } = \color{blue}{5x^{3}-22x^{2}-8x+77} \color{red}{~-~} \frac{ \color{red}{ 52 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-22&-8&77&-52\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 5 }&-22&-8&77&-52\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-22&-8&77&-52\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 0 } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}0&5&\color{orangered}{ -22 }&-8&77&-52\\& & \color{orangered}{0} & & & \\ \hline &5&\color{orangered}{-22}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-22&-8&77&-52\\& & 0& \color{blue}{0} & & \\ \hline &5&\color{blue}{-22}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}0&5&-22&\color{orangered}{ -8 }&77&-52\\& & 0& \color{orangered}{0} & & \\ \hline &5&-22&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-22&-8&77&-52\\& & 0& 0& \color{blue}{0} & \\ \hline &5&-22&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 77 } + \color{orangered}{ 0 } = \color{orangered}{ 77 } $
$$ \begin{array}{c|rrrrr}0&5&-22&-8&\color{orangered}{ 77 }&-52\\& & 0& 0& \color{orangered}{0} & \\ \hline &5&-22&-8&\color{orangered}{77}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 77 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&-22&-8&77&-52\\& & 0& 0& 0& \color{blue}{0} \\ \hline &5&-22&-8&\color{blue}{77}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 0 } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrrr}0&5&-22&-8&77&\color{orangered}{ -52 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{-22}&\color{blue}{-8}&\color{blue}{77}&\color{orangered}{-52} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-22x^{2}-8x+77 } $ with a remainder of $ \color{red}{ -52 } $.