Divide using synthetic division $$ ( 4x^{3}-6x^{2}+5x-1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&-6&5&-1\\& & 8& 4& \color{black}{18} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{17} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-6x^{2}+5x-1 }{ x-2 } = \color{blue}{4x^{2}+2x+9} ~+~ \frac{ \color{red}{ 17 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-6&5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-6&5&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-6&5&-1\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 8 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -6 }&5&-1\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-6&5&-1\\& & 8& \color{blue}{4} & \\ \hline &4&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 4 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}2&4&-6&\color{orangered}{ 5 }&-1\\& & 8& \color{orangered}{4} & \\ \hline &4&2&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-6&5&-1\\& & 8& 4& \color{blue}{18} \\ \hline &4&2&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 18 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}2&4&-6&5&\color{orangered}{ -1 }\\& & 8& 4& \color{orangered}{18} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{17} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+2x+9 } $ with a remainder of $ \color{red}{ 17 } $.