Divide using synthetic division $$ ( x^{4}+x^{3}-11x^{2}-5x+30 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&1&-11&-5&30\\& & 2& 6& -10& \color{black}{-30} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-5}&\color{blue}{-15}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}-11x^{2}-5x+30 }{ x-2 } = \color{blue}{x^{3}+3x^{2}-5x-15} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-11&-5&30\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&1&-11&-5&30\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-11&-5&30\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 1 }&-11&-5&30\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-11&-5&30\\& & 2& \color{blue}{6} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 6 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}2&1&1&\color{orangered}{ -11 }&-5&30\\& & 2& \color{orangered}{6} & & \\ \hline &1&3&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-11&-5&30\\& & 2& 6& \color{blue}{-10} & \\ \hline &1&3&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}2&1&1&-11&\color{orangered}{ -5 }&30\\& & 2& 6& \color{orangered}{-10} & \\ \hline &1&3&-5&\color{orangered}{-15}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-11&-5&30\\& & 2& 6& -10& \color{blue}{-30} \\ \hline &1&3&-5&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&1&1&-11&-5&\color{orangered}{ 30 }\\& & 2& 6& -10& \color{orangered}{-30} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-5}&\color{blue}{-15}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-5x-15 } $ with a remainder of $ \color{red}{ 0 } $.