Divide using synthetic division $$ ( 4x^{4}+8x^{3}-27x^{2}+17x-12 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&4&8&-27&17&-12\\& & -16& 32& -20& \color{black}{12} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+8x^{3}-27x^{2}+17x-12 }{ x+4 } = \color{blue}{4x^{3}-8x^{2}+5x-3} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&8&-27&17&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 4 }&8&-27&17&-12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&8&-27&17&-12\\& & \color{blue}{-16} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&4&\color{orangered}{ 8 }&-27&17&-12\\& & \color{orangered}{-16} & & & \\ \hline &4&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&8&-27&17&-12\\& & -16& \color{blue}{32} & & \\ \hline &4&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 32 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-4&4&8&\color{orangered}{ -27 }&17&-12\\& & -16& \color{orangered}{32} & & \\ \hline &4&-8&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&8&-27&17&-12\\& & -16& 32& \color{blue}{-20} & \\ \hline &4&-8&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&4&8&-27&\color{orangered}{ 17 }&-12\\& & -16& 32& \color{orangered}{-20} & \\ \hline &4&-8&5&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&8&-27&17&-12\\& & -16& 32& -20& \color{blue}{12} \\ \hline &4&-8&5&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&4&8&-27&17&\color{orangered}{ -12 }\\& & -16& 32& -20& \color{orangered}{12} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-8x^{2}+5x-3 } $ with a remainder of $ \color{red}{ 0 } $.