Divide using synthetic division $$ ( 4x^{5}+9x^{4}+4x^{2}+9x-14 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&4&9&0&4&9&-14\\& & -8& -2& 4& -16& \color{black}{14} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-2}&\color{blue}{8}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+9x^{4}+4x^{2}+9x-14 }{ x+2 } = \color{blue}{4x^{4}+x^{3}-2x^{2}+8x-7} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 4 }&9&0&4&9&-14\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & \color{blue}{-8} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-2&4&\color{orangered}{ 9 }&0&4&9&-14\\& & \color{orangered}{-8} & & & & \\ \hline &4&\color{orangered}{1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & -8& \color{blue}{-2} & & & \\ \hline &4&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&4&9&\color{orangered}{ 0 }&4&9&-14\\& & -8& \color{orangered}{-2} & & & \\ \hline &4&1&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & -8& -2& \color{blue}{4} & & \\ \hline &4&1&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 4 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}-2&4&9&0&\color{orangered}{ 4 }&9&-14\\& & -8& -2& \color{orangered}{4} & & \\ \hline &4&1&-2&\color{orangered}{8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & -8& -2& 4& \color{blue}{-16} & \\ \hline &4&1&-2&\color{blue}{8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-2&4&9&0&4&\color{orangered}{ 9 }&-14\\& & -8& -2& 4& \color{orangered}{-16} & \\ \hline &4&1&-2&8&\color{orangered}{-7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&4&9&0&4&9&-14\\& & -8& -2& 4& -16& \color{blue}{14} \\ \hline &4&1&-2&8&\color{blue}{-7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 14 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-2&4&9&0&4&9&\color{orangered}{ -14 }\\& & -8& -2& 4& -16& \color{orangered}{14} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-2}&\color{blue}{8}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+x^{3}-2x^{2}+8x-7 } $ with a remainder of $ \color{red}{ 0 } $.