Divide using synthetic division $$ ( -2x^{3}+3x-10 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&-2&0&3&-10\\& & 6& -18& \color{black}{45} \\ \hline &\color{blue}{-2}&\color{blue}{6}&\color{blue}{-15}&\color{orangered}{35} \end{array} $$The solution is:
$$ \frac{ -2x^{3}+3x-10 }{ x+3 } = \color{blue}{-2x^{2}+6x-15} ~+~ \frac{ \color{red}{ 35 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&0&3&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ -2 }&0&3&-10\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&0&3&-10\\& & \color{blue}{6} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-3&-2&\color{orangered}{ 0 }&3&-10\\& & \color{orangered}{6} & & \\ \hline &-2&\color{orangered}{6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&0&3&-10\\& & 6& \color{blue}{-18} & \\ \hline &-2&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-3&-2&0&\color{orangered}{ 3 }&-10\\& & 6& \color{orangered}{-18} & \\ \hline &-2&6&\color{orangered}{-15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&0&3&-10\\& & 6& -18& \color{blue}{45} \\ \hline &-2&6&\color{blue}{-15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 45 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrr}-3&-2&0&3&\color{orangered}{ -10 }\\& & 6& -18& \color{orangered}{45} \\ \hline &\color{blue}{-2}&\color{blue}{6}&\color{blue}{-15}&\color{orangered}{35} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{2}+6x-15 } $ with a remainder of $ \color{red}{ 35 } $.