Divide using synthetic division $$ ( 4x^{3}+10x^{2}+16x+40 ) \div ( -x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&4&10&16&40\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{16}&\color{orangered}{40} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+10x^{2}+16x+40 }{ x } = \color{blue}{4x^{2}+10x+16} ~+~ \frac{ \color{red}{ 40 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&10&16&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&10&16&40\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&10&16&40\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 0 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ 10 }&16&40\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 10 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&10&16&40\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 0 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}0&4&10&\color{orangered}{ 16 }&40\\& & 0& \color{orangered}{0} & \\ \hline &4&10&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 16 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&10&16&40\\& & 0& 0& \color{blue}{0} \\ \hline &4&10&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 0 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrr}0&4&10&16&\color{orangered}{ 40 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{16}&\color{orangered}{40} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+10x+16 } $ with a remainder of $ \color{red}{ 40 } $.