Divide using synthetic division $$ ( 16x^{4}+21x^{3}+10x^{2}+10x+2 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&16&21&10&10&2\\& & 48& 207& 651& \color{black}{1983} \\ \hline &\color{blue}{16}&\color{blue}{69}&\color{blue}{217}&\color{blue}{661}&\color{orangered}{1985} \end{array} $$The solution is:
$$ \frac{ 16x^{4}+21x^{3}+10x^{2}+10x+2 }{ x-3 } = \color{blue}{16x^{3}+69x^{2}+217x+661} ~+~ \frac{ \color{red}{ 1985 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&16&21&10&10&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 16 }&21&10&10&2\\& & & & & \\ \hline &\color{orangered}{16}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&16&21&10&10&2\\& & \color{blue}{48} & & & \\ \hline &\color{blue}{16}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 48 } = \color{orangered}{ 69 } $
$$ \begin{array}{c|rrrrr}3&16&\color{orangered}{ 21 }&10&10&2\\& & \color{orangered}{48} & & & \\ \hline &16&\color{orangered}{69}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 69 } = \color{blue}{ 207 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&16&21&10&10&2\\& & 48& \color{blue}{207} & & \\ \hline &16&\color{blue}{69}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 207 } = \color{orangered}{ 217 } $
$$ \begin{array}{c|rrrrr}3&16&21&\color{orangered}{ 10 }&10&2\\& & 48& \color{orangered}{207} & & \\ \hline &16&69&\color{orangered}{217}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 217 } = \color{blue}{ 651 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&16&21&10&10&2\\& & 48& 207& \color{blue}{651} & \\ \hline &16&69&\color{blue}{217}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 651 } = \color{orangered}{ 661 } $
$$ \begin{array}{c|rrrrr}3&16&21&10&\color{orangered}{ 10 }&2\\& & 48& 207& \color{orangered}{651} & \\ \hline &16&69&217&\color{orangered}{661}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 661 } = \color{blue}{ 1983 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&16&21&10&10&2\\& & 48& 207& 651& \color{blue}{1983} \\ \hline &16&69&217&\color{blue}{661}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1983 } = \color{orangered}{ 1985 } $
$$ \begin{array}{c|rrrrr}3&16&21&10&10&\color{orangered}{ 2 }\\& & 48& 207& 651& \color{orangered}{1983} \\ \hline &\color{blue}{16}&\color{blue}{69}&\color{blue}{217}&\color{blue}{661}&\color{orangered}{1985} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{3}+69x^{2}+217x+661 } $ with a remainder of $ \color{red}{ 1985 } $.