Divide using synthetic division $$ ( -5x^{4}+4x^{3}+2x^{2}-4 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&-5&4&2&0&-4\\& & 25& -145& 715& \color{black}{-3575} \\ \hline &\color{blue}{-5}&\color{blue}{29}&\color{blue}{-143}&\color{blue}{715}&\color{orangered}{-3579} \end{array} $$The solution is:
$$ \frac{ -5x^{4}+4x^{3}+2x^{2}-4 }{ x+5 } = \color{blue}{-5x^{3}+29x^{2}-143x+715} \color{red}{~-~} \frac{ \color{red}{ 3579 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&4&2&0&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ -5 }&4&2&0&-4\\& & & & & \\ \hline &\color{orangered}{-5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&4&2&0&-4\\& & \color{blue}{25} & & & \\ \hline &\color{blue}{-5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 25 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrrr}-5&-5&\color{orangered}{ 4 }&2&0&-4\\& & \color{orangered}{25} & & & \\ \hline &-5&\color{orangered}{29}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 29 } = \color{blue}{ -145 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&4&2&0&-4\\& & 25& \color{blue}{-145} & & \\ \hline &-5&\color{blue}{29}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -145 \right) } = \color{orangered}{ -143 } $
$$ \begin{array}{c|rrrrr}-5&-5&4&\color{orangered}{ 2 }&0&-4\\& & 25& \color{orangered}{-145} & & \\ \hline &-5&29&\color{orangered}{-143}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -143 \right) } = \color{blue}{ 715 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&4&2&0&-4\\& & 25& -145& \color{blue}{715} & \\ \hline &-5&29&\color{blue}{-143}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 715 } = \color{orangered}{ 715 } $
$$ \begin{array}{c|rrrrr}-5&-5&4&2&\color{orangered}{ 0 }&-4\\& & 25& -145& \color{orangered}{715} & \\ \hline &-5&29&-143&\color{orangered}{715}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 715 } = \color{blue}{ -3575 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-5&4&2&0&-4\\& & 25& -145& 715& \color{blue}{-3575} \\ \hline &-5&29&-143&\color{blue}{715}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -3575 \right) } = \color{orangered}{ -3579 } $
$$ \begin{array}{c|rrrrr}-5&-5&4&2&0&\color{orangered}{ -4 }\\& & 25& -145& 715& \color{orangered}{-3575} \\ \hline &\color{blue}{-5}&\color{blue}{29}&\color{blue}{-143}&\color{blue}{715}&\color{orangered}{-3579} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -5x^{3}+29x^{2}-143x+715 } $ with a remainder of $ \color{red}{ -3579 } $.