Divide using synthetic division $$ ( -2x^{3}-10x^{2}-13x-4 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&-2&-10&-13&-4\\& & 6& 12& \color{black}{3} \\ \hline &\color{blue}{-2}&\color{blue}{-4}&\color{blue}{-1}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ -2x^{3}-10x^{2}-13x-4 }{ x+3 } = \color{blue}{-2x^{2}-4x-1} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&-10&-13&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ -2 }&-10&-13&-4\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&-10&-13&-4\\& & \color{blue}{6} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&-2&\color{orangered}{ -10 }&-13&-4\\& & \color{orangered}{6} & & \\ \hline &-2&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&-10&-13&-4\\& & 6& \color{blue}{12} & \\ \hline &-2&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 12 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&-2&-10&\color{orangered}{ -13 }&-4\\& & 6& \color{orangered}{12} & \\ \hline &-2&-4&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-2&-10&-13&-4\\& & 6& 12& \color{blue}{3} \\ \hline &-2&-4&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 3 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&-2&-10&-13&\color{orangered}{ -4 }\\& & 6& 12& \color{orangered}{3} \\ \hline &\color{blue}{-2}&\color{blue}{-4}&\color{blue}{-1}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{2}-4x-1 } $ with a remainder of $ \color{red}{ -1 } $.