Divide using synthetic division $$ ( 4x^{5}+18x^{4}+7x^{2}-x-100 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&4&18&0&7&-1&-100\\& & -12& -18& 54& -183& \color{black}{552} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{-18}&\color{blue}{61}&\color{blue}{-184}&\color{orangered}{452} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+18x^{4}+7x^{2}-x-100 }{ x+3 } = \color{blue}{4x^{4}+6x^{3}-18x^{2}+61x-184} ~+~ \frac{ \color{red}{ 452 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 4 }&18&0&7&-1&-100\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & \color{blue}{-12} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-3&4&\color{orangered}{ 18 }&0&7&-1&-100\\& & \color{orangered}{-12} & & & & \\ \hline &4&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & -12& \color{blue}{-18} & & & \\ \hline &4&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrrr}-3&4&18&\color{orangered}{ 0 }&7&-1&-100\\& & -12& \color{orangered}{-18} & & & \\ \hline &4&6&\color{orangered}{-18}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & -12& -18& \color{blue}{54} & & \\ \hline &4&6&\color{blue}{-18}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 54 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrrrr}-3&4&18&0&\color{orangered}{ 7 }&-1&-100\\& & -12& -18& \color{orangered}{54} & & \\ \hline &4&6&-18&\color{orangered}{61}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 61 } = \color{blue}{ -183 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & -12& -18& 54& \color{blue}{-183} & \\ \hline &4&6&-18&\color{blue}{61}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -183 \right) } = \color{orangered}{ -184 } $
$$ \begin{array}{c|rrrrrr}-3&4&18&0&7&\color{orangered}{ -1 }&-100\\& & -12& -18& 54& \color{orangered}{-183} & \\ \hline &4&6&-18&61&\color{orangered}{-184}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -184 \right) } = \color{blue}{ 552 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&18&0&7&-1&-100\\& & -12& -18& 54& -183& \color{blue}{552} \\ \hline &4&6&-18&61&\color{blue}{-184}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -100 } + \color{orangered}{ 552 } = \color{orangered}{ 452 } $
$$ \begin{array}{c|rrrrrr}-3&4&18&0&7&-1&\color{orangered}{ -100 }\\& & -12& -18& 54& -183& \color{orangered}{552} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{-18}&\color{blue}{61}&\color{blue}{-184}&\color{orangered}{452} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+6x^{3}-18x^{2}+61x-184 } $ with a remainder of $ \color{red}{ 452 } $.