Divide using synthetic division $$ ( x^{3}+16x^{2}+63x+37 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&1&16&63&37\\& & -10& -60& \color{black}{-30} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{3}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{3}+16x^{2}+63x+37 }{ x+10 } = \color{blue}{x^{2}+6x+3} ~+~ \frac{ \color{red}{ 7 } }{ x+10 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&16&63&37\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 1 }&16&63&37\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&16&63&37\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-10&1&\color{orangered}{ 16 }&63&37\\& & \color{orangered}{-10} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 6 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&16&63&37\\& & -10& \color{blue}{-60} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 63 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-10&1&16&\color{orangered}{ 63 }&37\\& & -10& \color{orangered}{-60} & \\ \hline &1&6&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 3 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&16&63&37\\& & -10& -60& \color{blue}{-30} \\ \hline &1&6&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-10&1&16&63&\color{orangered}{ 37 }\\& & -10& -60& \color{orangered}{-30} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{3}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+6x+3 } $ with a remainder of $ \color{red}{ 7 } $.