Divide using synthetic division $$ ( 5x^{3}-4x^{2}+3x-2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&5&-4&3&-2\\& & 10& 12& \color{black}{30} \\ \hline &\color{blue}{5}&\color{blue}{6}&\color{blue}{15}&\color{orangered}{28} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-4x^{2}+3x-2 }{ x-2 } = \color{blue}{5x^{2}+6x+15} ~+~ \frac{ \color{red}{ 28 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-4&3&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 5 }&-4&3&-2\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-4&3&-2\\& & \color{blue}{10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 10 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}2&5&\color{orangered}{ -4 }&3&-2\\& & \color{orangered}{10} & & \\ \hline &5&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-4&3&-2\\& & 10& \color{blue}{12} & \\ \hline &5&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 12 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}2&5&-4&\color{orangered}{ 3 }&-2\\& & 10& \color{orangered}{12} & \\ \hline &5&6&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-4&3&-2\\& & 10& 12& \color{blue}{30} \\ \hline &5&6&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 30 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}2&5&-4&3&\color{orangered}{ -2 }\\& & 10& 12& \color{orangered}{30} \\ \hline &\color{blue}{5}&\color{blue}{6}&\color{blue}{15}&\color{orangered}{28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+6x+15 } $ with a remainder of $ \color{red}{ 28 } $.