Divide using synthetic division $$ ( 4x^{3}-21x^{2}+17x+12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&4&-21&17&12\\& & 16& -20& \color{black}{-12} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-21x^{2}+17x+12 }{ x-4 } = \color{blue}{4x^{2}-5x-3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&17&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 4 }&-21&17&12\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&17&12\\& & \color{blue}{16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 16 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&4&\color{orangered}{ -21 }&17&12\\& & \color{orangered}{16} & & \\ \hline &4&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&17&12\\& & 16& \color{blue}{-20} & \\ \hline &4&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}4&4&-21&\color{orangered}{ 17 }&12\\& & 16& \color{orangered}{-20} & \\ \hline &4&-5&\color{orangered}{-3}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&17&12\\& & 16& -20& \color{blue}{-12} \\ \hline &4&-5&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&4&-21&17&\color{orangered}{ 12 }\\& & 16& -20& \color{orangered}{-12} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-5x-3 } $ with a remainder of $ \color{red}{ 0 } $.