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