Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-7&-17&58&-24\\& & 8& 4& -52& \color{black}{24} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-13}&\color{blue}{6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-4 } = \color{blue}{2x^{3}+x^{2}-13x+6} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&-17&58&-24\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&-17&58&-24\\& & 8& \color{blue}{4} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 4 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}4&2&-7&\color{orangered}{ -17 }&58&-24\\& & 8& \color{orangered}{4} & & \\ \hline &2&1&\color{orangered}{-13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&-17&58&-24\\& & 8& 4& \color{blue}{-52} & \\ \hline &2&1&\color{blue}{-13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}4&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 8& 4& \color{orangered}{-52} & \\ \hline &2&1&-13&\color{orangered}{6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 6 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&-17&58&-24\\& & 8& 4& -52& \color{blue}{24} \\ \hline &2&1&-13&\color{blue}{6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 24 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&2&-7&-17&58&\color{orangered}{ -24 }\\& & 8& 4& -52& \color{orangered}{24} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-13}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}-13x+6 } $ with a remainder of $ \color{red}{ 0 } $.