Divide using synthetic division $$ ( 4x^{3}-15x^{2}+24x-17 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&-15&24&-17\\& & 8& -14& \color{black}{20} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{10}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-15x^{2}+24x-17 }{ x-2 } = \color{blue}{4x^{2}-7x+10} ~+~ \frac{ \color{red}{ 3 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-15&24&-17\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-15&24&-17\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-15&24&-17\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 8 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -15 }&24&-17\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-15&24&-17\\& & 8& \color{blue}{-14} & \\ \hline &4&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}2&4&-15&\color{orangered}{ 24 }&-17\\& & 8& \color{orangered}{-14} & \\ \hline &4&-7&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-15&24&-17\\& & 8& -14& \color{blue}{20} \\ \hline &4&-7&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 20 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}2&4&-15&24&\color{orangered}{ -17 }\\& & 8& -14& \color{orangered}{20} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{10}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-7x+10 } $ with a remainder of $ \color{red}{ 3 } $.