Divide using synthetic division $$ ( 4x^{3}-14x^{2}+17x-10 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&-14&17&-10\\& & 8& -12& \color{black}{10} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-14x^{2}+17x-10 }{ x-2 } = \color{blue}{4x^{2}-6x+5} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-14&17&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-14&17&-10\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-14&17&-10\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 8 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -14 }&17&-10\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-14&17&-10\\& & 8& \color{blue}{-12} & \\ \hline &4&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&4&-14&\color{orangered}{ 17 }&-10\\& & 8& \color{orangered}{-12} & \\ \hline &4&-6&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-14&17&-10\\& & 8& -12& \color{blue}{10} \\ \hline &4&-6&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&4&-14&17&\color{orangered}{ -10 }\\& & 8& -12& \color{orangered}{10} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-6x+5 } $ with a remainder of $ \color{red}{ 0 } $.