Divide using synthetic division $$ ( 3x^{4}-16x^{3}-7x^{2}+64x-20 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&3&-16&-7&64&-20\\& & 12& -16& -92& \color{black}{-112} \\ \hline &\color{blue}{3}&\color{blue}{-4}&\color{blue}{-23}&\color{blue}{-28}&\color{orangered}{-132} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-16x^{3}-7x^{2}+64x-20 }{ x-4 } = \color{blue}{3x^{3}-4x^{2}-23x-28} \color{red}{~-~} \frac{ \color{red}{ 132 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-16&-7&64&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 3 }&-16&-7&64&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-16&-7&64&-20\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 12 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&3&\color{orangered}{ -16 }&-7&64&-20\\& & \color{orangered}{12} & & & \\ \hline &3&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-16&-7&64&-20\\& & 12& \color{blue}{-16} & & \\ \hline &3&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrr}4&3&-16&\color{orangered}{ -7 }&64&-20\\& & 12& \color{orangered}{-16} & & \\ \hline &3&-4&\color{orangered}{-23}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ -92 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-16&-7&64&-20\\& & 12& -16& \color{blue}{-92} & \\ \hline &3&-4&\color{blue}{-23}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 64 } + \color{orangered}{ \left( -92 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrrr}4&3&-16&-7&\color{orangered}{ 64 }&-20\\& & 12& -16& \color{orangered}{-92} & \\ \hline &3&-4&-23&\color{orangered}{-28}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ -112 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-16&-7&64&-20\\& & 12& -16& -92& \color{blue}{-112} \\ \hline &3&-4&-23&\color{blue}{-28}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -112 \right) } = \color{orangered}{ -132 } $
$$ \begin{array}{c|rrrrr}4&3&-16&-7&64&\color{orangered}{ -20 }\\& & 12& -16& -92& \color{orangered}{-112} \\ \hline &\color{blue}{3}&\color{blue}{-4}&\color{blue}{-23}&\color{blue}{-28}&\color{orangered}{-132} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-4x^{2}-23x-28 } $ with a remainder of $ \color{red}{ -132 } $.