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