Divide using synthetic division $$ ( x^{5}-11x^{4}+10x^{3}+16x^{2} ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&-11&10&16&0&0\\& & 2& -18& -16& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-8}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}-11x^{4}+10x^{3}+16x^{2} }{ x-2 } = \color{blue}{x^{4}-9x^{3}-8x^{2}} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&-11&10&16&0&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 2 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ -11 }&10&16&0&0\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{-9}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & 2& \color{blue}{-18} & & & \\ \hline &1&\color{blue}{-9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}2&1&-11&\color{orangered}{ 10 }&16&0&0\\& & 2& \color{orangered}{-18} & & & \\ \hline &1&-9&\color{orangered}{-8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & 2& -18& \color{blue}{-16} & & \\ \hline &1&-9&\color{blue}{-8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-11&10&\color{orangered}{ 16 }&0&0\\& & 2& -18& \color{orangered}{-16} & & \\ \hline &1&-9&-8&\color{orangered}{0}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & 2& -18& -16& \color{blue}{0} & \\ \hline &1&-9&-8&\color{blue}{0}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-11&10&16&\color{orangered}{ 0 }&0\\& & 2& -18& -16& \color{orangered}{0} & \\ \hline &1&-9&-8&0&\color{orangered}{0}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-11&10&16&0&0\\& & 2& -18& -16& 0& \color{blue}{0} \\ \hline &1&-9&-8&0&\color{blue}{0}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-11&10&16&0&\color{orangered}{ 0 }\\& & 2& -18& -16& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-8}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-9x^{3}-8x^{2} } $ with a remainder of $ \color{red}{ 0 } $.