Divide using synthetic division $$ ( 3x^{3}-20x^{2}+12x-133 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&3&-20&12&-133\\& & 21& 7& \color{black}{133} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{19}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-20x^{2}+12x-133 }{ x-7 } = \color{blue}{3x^{2}+x+19} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&3&-20&12&-133\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 3 }&-20&12&-133\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 3 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&3&-20&12&-133\\& & \color{blue}{21} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 21 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}7&3&\color{orangered}{ -20 }&12&-133\\& & \color{orangered}{21} & & \\ \hline &3&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&3&-20&12&-133\\& & 21& \color{blue}{7} & \\ \hline &3&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 7 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}7&3&-20&\color{orangered}{ 12 }&-133\\& & 21& \color{orangered}{7} & \\ \hline &3&1&\color{orangered}{19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 19 } = \color{blue}{ 133 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&3&-20&12&-133\\& & 21& 7& \color{blue}{133} \\ \hline &3&1&\color{blue}{19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -133 } + \color{orangered}{ 133 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}7&3&-20&12&\color{orangered}{ -133 }\\& & 21& 7& \color{orangered}{133} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{19}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+x+19 } $ with a remainder of $ \color{red}{ 0 } $.