Divide using synthetic division $$ ( 2x^{4}+5x^{2}+15x-17 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&0&5&15&-17\\& & -8& 32& -148& \color{black}{532} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{37}&\color{blue}{-133}&\color{orangered}{515} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+5x^{2}+15x-17 }{ x+4 } = \color{blue}{2x^{3}-8x^{2}+37x-133} ~+~ \frac{ \color{red}{ 515 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&0&5&15&-17\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&0&5&15&-17\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&0&5&15&-17\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 0 }&5&15&-17\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&0&5&15&-17\\& & -8& \color{blue}{32} & & \\ \hline &2&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 32 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrrr}-4&2&0&\color{orangered}{ 5 }&15&-17\\& & -8& \color{orangered}{32} & & \\ \hline &2&-8&\color{orangered}{37}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 37 } = \color{blue}{ -148 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&0&5&15&-17\\& & -8& 32& \color{blue}{-148} & \\ \hline &2&-8&\color{blue}{37}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -148 \right) } = \color{orangered}{ -133 } $
$$ \begin{array}{c|rrrrr}-4&2&0&5&\color{orangered}{ 15 }&-17\\& & -8& 32& \color{orangered}{-148} & \\ \hline &2&-8&37&\color{orangered}{-133}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -133 \right) } = \color{blue}{ 532 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&0&5&15&-17\\& & -8& 32& -148& \color{blue}{532} \\ \hline &2&-8&37&\color{blue}{-133}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 532 } = \color{orangered}{ 515 } $
$$ \begin{array}{c|rrrrr}-4&2&0&5&15&\color{orangered}{ -17 }\\& & -8& 32& -148& \color{orangered}{532} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{37}&\color{blue}{-133}&\color{orangered}{515} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-8x^{2}+37x-133 } $ with a remainder of $ \color{red}{ 515 } $.