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