Divide using synthetic division $$ ( 2x^{4}-7x^{3}-17x^{2}+58x-24 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&2&-7&-17&58&-24\\& & 14& 49& 224& \color{black}{1974} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{32}&\color{blue}{282}&\color{orangered}{1950} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}-17x^{2}+58x-24 }{ x-7 } = \color{blue}{2x^{3}+7x^{2}+32x+282} ~+~ \frac{ \color{red}{ 1950 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&2&-7&-17&58&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 2 }&-7&-17&58&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 2 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&2&-7&-17&58&-24\\& & \color{blue}{14} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 14 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}7&2&\color{orangered}{ -7 }&-17&58&-24\\& & \color{orangered}{14} & & & \\ \hline &2&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 7 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&2&-7&-17&58&-24\\& & 14& \color{blue}{49} & & \\ \hline &2&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 49 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}7&2&-7&\color{orangered}{ -17 }&58&-24\\& & 14& \color{orangered}{49} & & \\ \hline &2&7&\color{orangered}{32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 32 } = \color{blue}{ 224 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&2&-7&-17&58&-24\\& & 14& 49& \color{blue}{224} & \\ \hline &2&7&\color{blue}{32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 58 } + \color{orangered}{ 224 } = \color{orangered}{ 282 } $
$$ \begin{array}{c|rrrrr}7&2&-7&-17&\color{orangered}{ 58 }&-24\\& & 14& 49& \color{orangered}{224} & \\ \hline &2&7&32&\color{orangered}{282}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 282 } = \color{blue}{ 1974 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&2&-7&-17&58&-24\\& & 14& 49& 224& \color{blue}{1974} \\ \hline &2&7&32&\color{blue}{282}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 1974 } = \color{orangered}{ 1950 } $
$$ \begin{array}{c|rrrrr}7&2&-7&-17&58&\color{orangered}{ -24 }\\& & 14& 49& 224& \color{orangered}{1974} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{32}&\color{blue}{282}&\color{orangered}{1950} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+7x^{2}+32x+282 } $ with a remainder of $ \color{red}{ 1950 } $.