Divide using synthetic division $$ ( 3x^{4}-12x+5 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&3&0&0&-12&5\\& & -3& 3& -3& \color{black}{15} \\ \hline &\color{blue}{3}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-15}&\color{orangered}{20} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-12x+5 }{ x+1 } = \color{blue}{3x^{3}-3x^{2}+3x-15} ~+~ \frac{ \color{red}{ 20 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&0&0&-12&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 3 }&0&0&-12&5\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&0&0&-12&5\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-1&3&\color{orangered}{ 0 }&0&-12&5\\& & \color{orangered}{-3} & & & \\ \hline &3&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&0&0&-12&5\\& & -3& \color{blue}{3} & & \\ \hline &3&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-1&3&0&\color{orangered}{ 0 }&-12&5\\& & -3& \color{orangered}{3} & & \\ \hline &3&-3&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&0&0&-12&5\\& & -3& 3& \color{blue}{-3} & \\ \hline &3&-3&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-1&3&0&0&\color{orangered}{ -12 }&5\\& & -3& 3& \color{orangered}{-3} & \\ \hline &3&-3&3&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&3&0&0&-12&5\\& & -3& 3& -3& \color{blue}{15} \\ \hline &3&-3&3&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 15 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-1&3&0&0&-12&\color{orangered}{ 5 }\\& & -3& 3& -3& \color{orangered}{15} \\ \hline &\color{blue}{3}&\color{blue}{-3}&\color{blue}{3}&\color{blue}{-15}&\color{orangered}{20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-3x^{2}+3x-15 } $ with a remainder of $ \color{red}{ 20 } $.