Divide using synthetic division $$ ( 5x^{3}-6x^{2}+8x-2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&5&-6&8&-2\\& & 10& 8& \color{black}{32} \\ \hline &\color{blue}{5}&\color{blue}{4}&\color{blue}{16}&\color{orangered}{30} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-6x^{2}+8x-2 }{ x-2 } = \color{blue}{5x^{2}+4x+16} ~+~ \frac{ \color{red}{ 30 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-6&8&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 5 }&-6&8&-2\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-6&8&-2\\& & \color{blue}{10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 10 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}2&5&\color{orangered}{ -6 }&8&-2\\& & \color{orangered}{10} & & \\ \hline &5&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-6&8&-2\\& & 10& \color{blue}{8} & \\ \hline &5&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 8 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}2&5&-6&\color{orangered}{ 8 }&-2\\& & 10& \color{orangered}{8} & \\ \hline &5&4&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 16 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&-6&8&-2\\& & 10& 8& \color{blue}{32} \\ \hline &5&4&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 32 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}2&5&-6&8&\color{orangered}{ -2 }\\& & 10& 8& \color{orangered}{32} \\ \hline &\color{blue}{5}&\color{blue}{4}&\color{blue}{16}&\color{orangered}{30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+4x+16 } $ with a remainder of $ \color{red}{ 30 } $.