Divide using synthetic division $$ ( 2x^{4}+5x^{3}-84x^{2}+37x+40 ) \div ( 1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&2&5&-84&37&40\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-84}&\color{blue}{37}&\color{orangered}{40} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+5x^{3}-84x^{2}+37x+40 }{ x } = \color{blue}{2x^{3}+5x^{2}-84x+37} ~+~ \frac{ \color{red}{ 40 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&5&-84&37&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 2 }&5&-84&37&40\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&5&-84&37&40\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}0&2&\color{orangered}{ 5 }&-84&37&40\\& & \color{orangered}{0} & & & \\ \hline &2&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&5&-84&37&40\\& & 0& \color{blue}{0} & & \\ \hline &2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -84 } + \color{orangered}{ 0 } = \color{orangered}{ -84 } $
$$ \begin{array}{c|rrrrr}0&2&5&\color{orangered}{ -84 }&37&40\\& & 0& \color{orangered}{0} & & \\ \hline &2&5&\color{orangered}{-84}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -84 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&5&-84&37&40\\& & 0& 0& \color{blue}{0} & \\ \hline &2&5&\color{blue}{-84}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ 0 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrrr}0&2&5&-84&\color{orangered}{ 37 }&40\\& & 0& 0& \color{orangered}{0} & \\ \hline &2&5&-84&\color{orangered}{37}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 37 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&5&-84&37&40\\& & 0& 0& 0& \color{blue}{0} \\ \hline &2&5&-84&\color{blue}{37}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 0 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrrr}0&2&5&-84&37&\color{orangered}{ 40 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-84}&\color{blue}{37}&\color{orangered}{40} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+5x^{2}-84x+37 } $ with a remainder of $ \color{red}{ 40 } $.