Check if $ x-4 $ is a factor of $ 5x^{4}-22x^{3}-8x^{2}+77x-52 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&5&-22&-8&77&-52\\& & 20& -8& -64& \color{black}{52} \\ \hline &\color{blue}{5}&\color{blue}{-2}&\color{blue}{-16}&\color{blue}{13}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-4 $ is a factor of the $ 5x^{4}-22x^{3}-8x^{2}+77x-52 $.
Explanation
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-22&-8&77&-52\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 5 }&-22&-8&77&-52\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-22&-8&77&-52\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 20 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&5&\color{orangered}{ -22 }&-8&77&-52\\& & \color{orangered}{20} & & & \\ \hline &5&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-22&-8&77&-52\\& & 20& \color{blue}{-8} & & \\ \hline &5&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}4&5&-22&\color{orangered}{ -8 }&77&-52\\& & 20& \color{orangered}{-8} & & \\ \hline &5&-2&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-22&-8&77&-52\\& & 20& -8& \color{blue}{-64} & \\ \hline &5&-2&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 77 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}4&5&-22&-8&\color{orangered}{ 77 }&-52\\& & 20& -8& \color{orangered}{-64} & \\ \hline &5&-2&-16&\color{orangered}{13}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 13 } = \color{blue}{ 52 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-22&-8&77&-52\\& & 20& -8& -64& \color{blue}{52} \\ \hline &5&-2&-16&\color{blue}{13}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 52 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&5&-22&-8&77&\color{orangered}{ -52 }\\& & 20& -8& -64& \color{orangered}{52} \\ \hline &\color{blue}{5}&\color{blue}{-2}&\color{blue}{-16}&\color{blue}{13}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.