Find remainder when $ -28x^{4}-7x^{3}-16x^{2}+5x $ is divided by $ -5 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&-28&-7&-16&5&0\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{-28}&\color{blue}{-7}&\color{blue}{-16}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The remainder when $ -28x^{4}-7x^{3}-16x^{2}+5x $ is divided by $ x $ is $ \, \color{red}{ 0 } $.
We can find remainder using synthetic division method.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&-28&-7&-16&5&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ -28 }&-7&-16&5&0\\& & & & & \\ \hline &\color{orangered}{-28}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&-28&-7&-16&5&0\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{-28}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}0&-28&\color{orangered}{ -7 }&-16&5&0\\& & \color{orangered}{0} & & & \\ \hline &-28&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&-28&-7&-16&5&0\\& & 0& \color{blue}{0} & & \\ \hline &-28&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 0 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}0&-28&-7&\color{orangered}{ -16 }&5&0\\& & 0& \color{orangered}{0} & & \\ \hline &-28&-7&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&-28&-7&-16&5&0\\& & 0& 0& \color{blue}{0} & \\ \hline &-28&-7&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}0&-28&-7&-16&\color{orangered}{ 5 }&0\\& & 0& 0& \color{orangered}{0} & \\ \hline &-28&-7&-16&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&-28&-7&-16&5&0\\& & 0& 0& 0& \color{blue}{0} \\ \hline &-28&-7&-16&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}0&-28&-7&-16&5&\color{orangered}{ 0 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{-28}&\color{blue}{-7}&\color{blue}{-16}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.