Divide using synthetic division $$ ( x^{3}-8x^{2}+19x-12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-8&19&-12\\& & 3& -15& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-8x^{2}+19x-12 }{ x-3 } = \color{blue}{x^{2}-5x+4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-8&19&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-8&19&-12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-8&19&-12\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 3 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -8 }&19&-12\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-8&19&-12\\& & 3& \color{blue}{-15} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}3&1&-8&\color{orangered}{ 19 }&-12\\& & 3& \color{orangered}{-15} & \\ \hline &1&-5&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-8&19&-12\\& & 3& -15& \color{blue}{12} \\ \hline &1&-5&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&-8&19&\color{orangered}{ -12 }\\& & 3& -15& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+4 } $ with a remainder of $ \color{red}{ 0 } $.