Divide using synthetic division $$ ( 6x^{3}-5x^{2}+2x-5 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&6&-5&2&-5\\& & 12& 14& \color{black}{32} \\ \hline &\color{blue}{6}&\color{blue}{7}&\color{blue}{16}&\color{orangered}{27} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-5x^{2}+2x-5 }{ x-2 } = \color{blue}{6x^{2}+7x+16} ~+~ \frac{ \color{red}{ 27 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-5&2&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 6 }&-5&2&-5\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-5&2&-5\\& & \color{blue}{12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 12 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}2&6&\color{orangered}{ -5 }&2&-5\\& & \color{orangered}{12} & & \\ \hline &6&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-5&2&-5\\& & 12& \color{blue}{14} & \\ \hline &6&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 14 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}2&6&-5&\color{orangered}{ 2 }&-5\\& & 12& \color{orangered}{14} & \\ \hline &6&7&\color{orangered}{16}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-5&2&-5\\& & 12& 14& \color{blue}{32} \\ \hline &6&7&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 32 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrr}2&6&-5&2&\color{orangered}{ -5 }\\& & 12& 14& \color{orangered}{32} \\ \hline &\color{blue}{6}&\color{blue}{7}&\color{blue}{16}&\color{orangered}{27} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+7x+16 } $ with a remainder of $ \color{red}{ 27 } $.