Divide using synthetic division $$ ( 3x^{4}+10x^{3}-7x^{2}+4x ) \div ( -x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&3&10&-7&4&0\\& & 12& 88& 324& \color{black}{1312} \\ \hline &\color{blue}{3}&\color{blue}{22}&\color{blue}{81}&\color{blue}{328}&\color{orangered}{1312} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+10x^{3}-7x^{2}+4x }{ x-4 } = \color{blue}{3x^{3}+22x^{2}+81x+328} ~+~ \frac{ \color{red}{ 1312 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&10&-7&4&0\\& & & & & \\ \hline &&&&& \end{array} $$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}{ 12 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}4&3&\color{orangered}{ 10 }&-7&4&0\\& & \color{orangered}{12} & & & \\ \hline &3&\color{orangered}{22}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 22 } = \color{blue}{ 88 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&10&-7&4&0\\& & 12& \color{blue}{88} & & \\ \hline &3&\color{blue}{22}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 88 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrr}4&3&10&\color{orangered}{ -7 }&4&0\\& & 12& \color{orangered}{88} & & \\ \hline &3&22&\color{orangered}{81}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 81 } = \color{blue}{ 324 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&10&-7&4&0\\& & 12& 88& \color{blue}{324} & \\ \hline &3&22&\color{blue}{81}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 324 } = \color{orangered}{ 328 } $
$$ \begin{array}{c|rrrrr}4&3&10&-7&\color{orangered}{ 4 }&0\\& & 12& 88& \color{orangered}{324} & \\ \hline &3&22&81&\color{orangered}{328}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 328 } = \color{blue}{ 1312 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&10&-7&4&0\\& & 12& 88& 324& \color{blue}{1312} \\ \hline &3&22&81&\color{blue}{328}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1312 } = \color{orangered}{ 1312 } $
$$ \begin{array}{c|rrrrr}4&3&10&-7&4&\color{orangered}{ 0 }\\& & 12& 88& 324& \color{orangered}{1312} \\ \hline &\color{blue}{3}&\color{blue}{22}&\color{blue}{81}&\color{blue}{328}&\color{orangered}{1312} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+22x^{2}+81x+328 } $ with a remainder of $ \color{red}{ 1312 } $.