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