Divide using synthetic division $$ ( x^{4}-2x^{3}-16x^{2}+28x+9 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&-2&-16&28&9\\& & 4& 8& -32& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{-4}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-16x^{2}+28x+9 }{ x-4 } = \color{blue}{x^{3}+2x^{2}-8x-4} \color{red}{~-~} \frac{ \color{red}{ 7 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-2&-16&28&9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&-2&-16&28&9\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-2&-16&28&9\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ -2 }&-16&28&9\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-2&-16&28&9\\& & 4& \color{blue}{8} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 8 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}4&1&-2&\color{orangered}{ -16 }&28&9\\& & 4& \color{orangered}{8} & & \\ \hline &1&2&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-2&-16&28&9\\& & 4& 8& \color{blue}{-32} & \\ \hline &1&2&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&1&-2&-16&\color{orangered}{ 28 }&9\\& & 4& 8& \color{orangered}{-32} & \\ \hline &1&2&-8&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-2&-16&28&9\\& & 4& 8& -32& \color{blue}{-16} \\ \hline &1&2&-8&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}4&1&-2&-16&28&\color{orangered}{ 9 }\\& & 4& 8& -32& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{-4}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-8x-4 } $ with a remainder of $ \color{red}{ -7 } $.