Divide using synthetic division $$ ( 2x^{4}-15x^{2}-16x-4 ) \div ( x+52 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-52&2&0&-15&-16&-4\\& & -104& 5408& -280436& \color{black}{14583504} \\ \hline &\color{blue}{2}&\color{blue}{-104}&\color{blue}{5393}&\color{blue}{-280452}&\color{orangered}{14583500} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-15x^{2}-16x-4 }{ x+52 } = \color{blue}{2x^{3}-104x^{2}+5393x-280452} ~+~ \frac{ \color{red}{ 14583500 } }{ x+52 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 52 = 0 $ ( $ x = \color{blue}{ -52 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-52}&2&0&-15&-16&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-52&\color{orangered}{ 2 }&0&-15&-16&-4\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -52 } \cdot \color{blue}{ 2 } = \color{blue}{ -104 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-52}&2&0&-15&-16&-4\\& & \color{blue}{-104} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -104 \right) } = \color{orangered}{ -104 } $
$$ \begin{array}{c|rrrrr}-52&2&\color{orangered}{ 0 }&-15&-16&-4\\& & \color{orangered}{-104} & & & \\ \hline &2&\color{orangered}{-104}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -52 } \cdot \color{blue}{ \left( -104 \right) } = \color{blue}{ 5408 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-52}&2&0&-15&-16&-4\\& & -104& \color{blue}{5408} & & \\ \hline &2&\color{blue}{-104}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 5408 } = \color{orangered}{ 5393 } $
$$ \begin{array}{c|rrrrr}-52&2&0&\color{orangered}{ -15 }&-16&-4\\& & -104& \color{orangered}{5408} & & \\ \hline &2&-104&\color{orangered}{5393}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -52 } \cdot \color{blue}{ 5393 } = \color{blue}{ -280436 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-52}&2&0&-15&-16&-4\\& & -104& 5408& \color{blue}{-280436} & \\ \hline &2&-104&\color{blue}{5393}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -280436 \right) } = \color{orangered}{ -280452 } $
$$ \begin{array}{c|rrrrr}-52&2&0&-15&\color{orangered}{ -16 }&-4\\& & -104& 5408& \color{orangered}{-280436} & \\ \hline &2&-104&5393&\color{orangered}{-280452}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -52 } \cdot \color{blue}{ \left( -280452 \right) } = \color{blue}{ 14583504 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-52}&2&0&-15&-16&-4\\& & -104& 5408& -280436& \color{blue}{14583504} \\ \hline &2&-104&5393&\color{blue}{-280452}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 14583504 } = \color{orangered}{ 14583500 } $
$$ \begin{array}{c|rrrrr}-52&2&0&-15&-16&\color{orangered}{ -4 }\\& & -104& 5408& -280436& \color{orangered}{14583504} \\ \hline &\color{blue}{2}&\color{blue}{-104}&\color{blue}{5393}&\color{blue}{-280452}&\color{orangered}{14583500} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-104x^{2}+5393x-280452 } $ with a remainder of $ \color{red}{ 14583500 } $.