Divide using synthetic division $$ ( 2x^{5}+3x^{4}+25x^{2}-1 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&2&3&0&25&0&-1\\& & -6& 9& -27& 6& \color{black}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{-19} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+3x^{4}+25x^{2}-1 }{ x+3 } = \color{blue}{2x^{4}-3x^{3}+9x^{2}-2x+6} \color{red}{~-~} \frac{ \color{red}{ 19 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 2 }&3&0&25&0&-1\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & \color{blue}{-6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&2&\color{orangered}{ 3 }&0&25&0&-1\\& & \color{orangered}{-6} & & & & \\ \hline &2&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & -6& \color{blue}{9} & & & \\ \hline &2&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-3&2&3&\color{orangered}{ 0 }&25&0&-1\\& & -6& \color{orangered}{9} & & & \\ \hline &2&-3&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 9 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & -6& 9& \color{blue}{-27} & & \\ \hline &2&-3&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-3&2&3&0&\color{orangered}{ 25 }&0&-1\\& & -6& 9& \color{orangered}{-27} & & \\ \hline &2&-3&9&\color{orangered}{-2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & -6& 9& -27& \color{blue}{6} & \\ \hline &2&-3&9&\color{blue}{-2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-3&2&3&0&25&\color{orangered}{ 0 }&-1\\& & -6& 9& -27& \color{orangered}{6} & \\ \hline &2&-3&9&-2&\color{orangered}{6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&2&3&0&25&0&-1\\& & -6& 9& -27& 6& \color{blue}{-18} \\ \hline &2&-3&9&-2&\color{blue}{6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrrrr}-3&2&3&0&25&0&\color{orangered}{ -1 }\\& & -6& 9& -27& 6& \color{orangered}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{6}&\color{orangered}{-19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-3x^{3}+9x^{2}-2x+6 } $ with a remainder of $ \color{red}{ -19 } $.