Divide using synthetic division $$ ( -11x^{4}+10x^{3}+16x^{2}+x ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&-11&10&16&1&0\\& & -22& -24& -16& \color{black}{-30} \\ \hline &\color{blue}{-11}&\color{blue}{-12}&\color{blue}{-8}&\color{blue}{-15}&\color{orangered}{-30} \end{array} $$The solution is:
$$ \frac{ -11x^{4}+10x^{3}+16x^{2}+x }{ x-2 } = \color{blue}{-11x^{3}-12x^{2}-8x-15} \color{red}{~-~} \frac{ \color{red}{ 30 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-11&10&16&1&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ -11 }&10&16&1&0\\& & & & & \\ \hline &\color{orangered}{-11}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-11&10&16&1&0\\& & \color{blue}{-22} & & & \\ \hline &\color{blue}{-11}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}2&-11&\color{orangered}{ 10 }&16&1&0\\& & \color{orangered}{-22} & & & \\ \hline &-11&\color{orangered}{-12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-11&10&16&1&0\\& & -22& \color{blue}{-24} & & \\ \hline &-11&\color{blue}{-12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}2&-11&10&\color{orangered}{ 16 }&1&0\\& & -22& \color{orangered}{-24} & & \\ \hline &-11&-12&\color{orangered}{-8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-11&10&16&1&0\\& & -22& -24& \color{blue}{-16} & \\ \hline &-11&-12&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}2&-11&10&16&\color{orangered}{ 1 }&0\\& & -22& -24& \color{orangered}{-16} & \\ \hline &-11&-12&-8&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-11&10&16&1&0\\& & -22& -24& -16& \color{blue}{-30} \\ \hline &-11&-12&-8&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrrr}2&-11&10&16&1&\color{orangered}{ 0 }\\& & -22& -24& -16& \color{orangered}{-30} \\ \hline &\color{blue}{-11}&\color{blue}{-12}&\color{blue}{-8}&\color{blue}{-15}&\color{orangered}{-30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -11x^{3}-12x^{2}-8x-15 } $ with a remainder of $ \color{red}{ -30 } $.