Divide using synthetic division $$ ( -2x^{4}-x^{2}+3 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&-2&0&-1&0&3\\& & 6& -18& 57& \color{black}{-171} \\ \hline &\color{blue}{-2}&\color{blue}{6}&\color{blue}{-19}&\color{blue}{57}&\color{orangered}{-168} \end{array} $$The solution is:
$$ \frac{ -2x^{4}-x^{2}+3 }{ x+3 } = \color{blue}{-2x^{3}+6x^{2}-19x+57} \color{red}{~-~} \frac{ \color{red}{ 168 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-2&0&-1&0&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ -2 }&0&-1&0&3\\& & & & & \\ \hline &\color{orangered}{-2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-2&0&-1&0&3\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{-2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-3&-2&\color{orangered}{ 0 }&-1&0&3\\& & \color{orangered}{6} & & & \\ \hline &-2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-2&0&-1&0&3\\& & 6& \color{blue}{-18} & & \\ \hline &-2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrrr}-3&-2&0&\color{orangered}{ -1 }&0&3\\& & 6& \color{orangered}{-18} & & \\ \hline &-2&6&\color{orangered}{-19}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-2&0&-1&0&3\\& & 6& -18& \color{blue}{57} & \\ \hline &-2&6&\color{blue}{-19}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 57 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrrr}-3&-2&0&-1&\color{orangered}{ 0 }&3\\& & 6& -18& \color{orangered}{57} & \\ \hline &-2&6&-19&\color{orangered}{57}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 57 } = \color{blue}{ -171 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-2&0&-1&0&3\\& & 6& -18& 57& \color{blue}{-171} \\ \hline &-2&6&-19&\color{blue}{57}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -171 \right) } = \color{orangered}{ -168 } $
$$ \begin{array}{c|rrrrr}-3&-2&0&-1&0&\color{orangered}{ 3 }\\& & 6& -18& 57& \color{orangered}{-171} \\ \hline &\color{blue}{-2}&\color{blue}{6}&\color{blue}{-19}&\color{blue}{57}&\color{orangered}{-168} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{3}+6x^{2}-19x+57 } $ with a remainder of $ \color{red}{ -168 } $.