Divide using synthetic division $$ ( 3x^{4}-16x^{3}-7x^{2}+64x-20 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&-16&-7&64&-20\\& & 6& -20& -54& \color{black}{20} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{-27}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-16x^{3}-7x^{2}+64x-20 }{ x-2 } = \color{blue}{3x^{3}-10x^{2}-27x+10} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-16&-7&64&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&-16&-7&64&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-16&-7&64&-20\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 6 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-16&-7&64&-20\\& & 6& \color{blue}{-20} & & \\ \hline &3&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrrr}2&3&-16&\color{orangered}{ -7 }&64&-20\\& & 6& \color{orangered}{-20} & & \\ \hline &3&-10&\color{orangered}{-27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-16&-7&64&-20\\& & 6& -20& \color{blue}{-54} & \\ \hline &3&-10&\color{blue}{-27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&3&-16&-7&\color{orangered}{ 64 }&-20\\& & 6& -20& \color{orangered}{-54} & \\ \hline &3&-10&-27&\color{orangered}{10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-16&-7&64&-20\\& & 6& -20& -54& \color{blue}{20} \\ \hline &3&-10&-27&\color{blue}{10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&3&-16&-7&64&\color{orangered}{ -20 }\\& & 6& -20& -54& \color{orangered}{20} \\ \hline &\color{blue}{3}&\color{blue}{-10}&\color{blue}{-27}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-10x^{2}-27x+10 } $ with a remainder of $ \color{red}{ 0 } $.