Divide using synthetic division $$ ( 3x^{4}+16x^{3}+4x^{2}-5x ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&3&16&4&-5&0\\& & -15& -5& 5& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-1}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+16x^{3}+4x^{2}-5x }{ x+5 } = \color{blue}{3x^{3}+x^{2}-x} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&16&4&-5&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 3 }&16&4&-5&0\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&16&4&-5&0\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-5&3&\color{orangered}{ 16 }&4&-5&0\\& & \color{orangered}{-15} & & & \\ \hline &3&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&16&4&-5&0\\& & -15& \color{blue}{-5} & & \\ \hline &3&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-5&3&16&\color{orangered}{ 4 }&-5&0\\& & -15& \color{orangered}{-5} & & \\ \hline &3&1&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&16&4&-5&0\\& & -15& -5& \color{blue}{5} & \\ \hline &3&1&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 5 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&3&16&4&\color{orangered}{ -5 }&0\\& & -15& -5& \color{orangered}{5} & \\ \hline &3&1&-1&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&16&4&-5&0\\& & -15& -5& 5& \color{blue}{0} \\ \hline &3&1&-1&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&3&16&4&-5&\color{orangered}{ 0 }\\& & -15& -5& 5& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-1}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+x^{2}-x } $ with a remainder of $ \color{red}{ 0 } $.