Divide using synthetic division $$ ( 7x^{4}+30x^{3}-30x-10 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&7&30&0&-30&-10\\& & -28& -8& 32& \color{black}{-8} \\ \hline &\color{blue}{7}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{2}&\color{orangered}{-18} \end{array} $$The solution is:
$$ \frac{ 7x^{4}+30x^{3}-30x-10 }{ x+4 } = \color{blue}{7x^{3}+2x^{2}-8x+2} \color{red}{~-~} \frac{ \color{red}{ 18 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&30&0&-30&-10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 7 }&30&0&-30&-10\\& & & & & \\ \hline &\color{orangered}{7}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 7 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&30&0&-30&-10\\& & \color{blue}{-28} & & & \\ \hline &\color{blue}{7}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-4&7&\color{orangered}{ 30 }&0&-30&-10\\& & \color{orangered}{-28} & & & \\ \hline &7&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&30&0&-30&-10\\& & -28& \color{blue}{-8} & & \\ \hline &7&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&7&30&\color{orangered}{ 0 }&-30&-10\\& & -28& \color{orangered}{-8} & & \\ \hline &7&2&\color{orangered}{-8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&30&0&-30&-10\\& & -28& -8& \color{blue}{32} & \\ \hline &7&2&\color{blue}{-8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 32 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-4&7&30&0&\color{orangered}{ -30 }&-10\\& & -28& -8& \color{orangered}{32} & \\ \hline &7&2&-8&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&30&0&-30&-10\\& & -28& -8& 32& \color{blue}{-8} \\ \hline &7&2&-8&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrr}-4&7&30&0&-30&\color{orangered}{ -10 }\\& & -28& -8& 32& \color{orangered}{-8} \\ \hline &\color{blue}{7}&\color{blue}{2}&\color{blue}{-8}&\color{blue}{2}&\color{orangered}{-18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{3}+2x^{2}-8x+2 } $ with a remainder of $ \color{red}{ -18 } $.