Divide using synthetic division $$ ( 2x^{4}+9x^{3}-5x+2 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&9&0&-5&2\\& & -8& -4& 16& \color{black}{-44} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{11}&\color{orangered}{-42} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+9x^{3}-5x+2 }{ x+4 } = \color{blue}{2x^{3}+x^{2}-4x+11} \color{red}{~-~} \frac{ \color{red}{ 42 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&9&0&-5&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&9&0&-5&2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&9&0&-5&2\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 9 }&0&-5&2\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&9&0&-5&2\\& & -8& \color{blue}{-4} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&2&9&\color{orangered}{ 0 }&-5&2\\& & -8& \color{orangered}{-4} & & \\ \hline &2&1&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&9&0&-5&2\\& & -8& -4& \color{blue}{16} & \\ \hline &2&1&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 16 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}-4&2&9&0&\color{orangered}{ -5 }&2\\& & -8& -4& \color{orangered}{16} & \\ \hline &2&1&-4&\color{orangered}{11}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 11 } = \color{blue}{ -44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&9&0&-5&2\\& & -8& -4& 16& \color{blue}{-44} \\ \hline &2&1&-4&\color{blue}{11}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -44 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrr}-4&2&9&0&-5&\color{orangered}{ 2 }\\& & -8& -4& 16& \color{orangered}{-44} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-4}&\color{blue}{11}&\color{orangered}{-42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}-4x+11 } $ with a remainder of $ \color{red}{ -42 } $.