Divide using synthetic division $$ ( 4x^{4}+52x^{3}+87x^{2}-13x-22 ) \div ( -2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&4&52&87&-13&-22\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{52}&\color{blue}{87}&\color{blue}{-13}&\color{orangered}{-22} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+52x^{3}+87x^{2}-13x-22 }{ x } = \color{blue}{4x^{3}+52x^{2}+87x-13} \color{red}{~-~} \frac{ \color{red}{ 22 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&52&87&-13&-22\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 4 }&52&87&-13&-22\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&52&87&-13&-22\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ 0 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrrr}0&4&\color{orangered}{ 52 }&87&-13&-22\\& & \color{orangered}{0} & & & \\ \hline &4&\color{orangered}{52}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 52 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&52&87&-13&-22\\& & 0& \color{blue}{0} & & \\ \hline &4&\color{blue}{52}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 87 } + \color{orangered}{ 0 } = \color{orangered}{ 87 } $
$$ \begin{array}{c|rrrrr}0&4&52&\color{orangered}{ 87 }&-13&-22\\& & 0& \color{orangered}{0} & & \\ \hline &4&52&\color{orangered}{87}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 87 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&52&87&-13&-22\\& & 0& 0& \color{blue}{0} & \\ \hline &4&52&\color{blue}{87}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 0 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}0&4&52&87&\color{orangered}{ -13 }&-22\\& & 0& 0& \color{orangered}{0} & \\ \hline &4&52&87&\color{orangered}{-13}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&4&52&87&-13&-22\\& & 0& 0& 0& \color{blue}{0} \\ \hline &4&52&87&\color{blue}{-13}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 0 } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}0&4&52&87&-13&\color{orangered}{ -22 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{52}&\color{blue}{87}&\color{blue}{-13}&\color{orangered}{-22} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+52x^{2}+87x-13 } $ with a remainder of $ \color{red}{ -22 } $.