Divide using synthetic division $$ ( -4x^{4}-10x^{3}+5x^{2}+7 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&-4&-10&5&0&7\\& & 12& -6& 3& \color{black}{-9} \\ \hline &\color{blue}{-4}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ -4x^{4}-10x^{3}+5x^{2}+7 }{ x+3 } = \color{blue}{-4x^{3}+2x^{2}-x+3} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-4&-10&5&0&7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ -4 }&-10&5&0&7\\& & & & & \\ \hline &\color{orangered}{-4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-4&-10&5&0&7\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{-4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&-4&\color{orangered}{ -10 }&5&0&7\\& & \color{orangered}{12} & & & \\ \hline &-4&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-4&-10&5&0&7\\& & 12& \color{blue}{-6} & & \\ \hline &-4&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&-4&-10&\color{orangered}{ 5 }&0&7\\& & 12& \color{orangered}{-6} & & \\ \hline &-4&2&\color{orangered}{-1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-4&-10&5&0&7\\& & 12& -6& \color{blue}{3} & \\ \hline &-4&2&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&-4&-10&5&\color{orangered}{ 0 }&7\\& & 12& -6& \color{orangered}{3} & \\ \hline &-4&2&-1&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&-4&-10&5&0&7\\& & 12& -6& 3& \color{blue}{-9} \\ \hline &-4&2&-1&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&-4&-10&5&0&\color{orangered}{ 7 }\\& & 12& -6& 3& \color{orangered}{-9} \\ \hline &\color{blue}{-4}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{3}+2x^{2}-x+3 } $ with a remainder of $ \color{red}{ -2 } $.