Divide using synthetic division $$ ( 4x^{5}+x^{4}+4x^{2}+5 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&4&1&0&4&0&5\\& & -8& 14& -28& 48& \color{black}{-96} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{14}&\color{blue}{-24}&\color{blue}{48}&\color{orangered}{-91} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+x^{4}+4x^{2}+5 }{ x+2 } = \color{blue}{4x^{4}-7x^{3}+14x^{2}-24x+48} \color{red}{~-~} \frac{ \color{red}{ 91 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 4 }&1&0&4&0&5\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & \color{blue}{-8} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-2&4&\color{orangered}{ 1 }&0&4&0&5\\& & \color{orangered}{-8} & & & & \\ \hline &4&\color{orangered}{-7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & -8& \color{blue}{14} & & & \\ \hline &4&\color{blue}{-7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 14 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrrr}-2&4&1&\color{orangered}{ 0 }&4&0&5\\& & -8& \color{orangered}{14} & & & \\ \hline &4&-7&\color{orangered}{14}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 14 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & -8& 14& \color{blue}{-28} & & \\ \hline &4&-7&\color{blue}{14}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrrr}-2&4&1&0&\color{orangered}{ 4 }&0&5\\& & -8& 14& \color{orangered}{-28} & & \\ \hline &4&-7&14&\color{orangered}{-24}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & -8& 14& -28& \color{blue}{48} & \\ \hline &4&-7&14&\color{blue}{-24}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 48 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrrr}-2&4&1&0&4&\color{orangered}{ 0 }&5\\& & -8& 14& -28& \color{orangered}{48} & \\ \hline &4&-7&14&-24&\color{orangered}{48}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 48 } = \color{blue}{ -96 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&1&0&4&0&5\\& & -8& 14& -28& 48& \color{blue}{-96} \\ \hline &4&-7&14&-24&\color{blue}{48}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -96 \right) } = \color{orangered}{ -91 } $
$$ \begin{array}{c|rrrrrr}-2&4&1&0&4&0&\color{orangered}{ 5 }\\& & -8& 14& -28& 48& \color{orangered}{-96} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{14}&\color{blue}{-24}&\color{blue}{48}&\color{orangered}{-91} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}-7x^{3}+14x^{2}-24x+48 } $ with a remainder of $ \color{red}{ -91 } $.