Divide using synthetic division $$ ( 2x^{4}-7x^{3}+8x+2 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-7&0&8&2\\& & 8& 4& 16& \color{black}{96} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{4}&\color{blue}{24}&\color{orangered}{98} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{3}+8x+2 }{ x-4 } = \color{blue}{2x^{3}+x^{2}+4x+24} ~+~ \frac{ \color{red}{ 98 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&0&8&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-7&0&8&2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&0&8&2\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -7 }&0&8&2\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&0&8&2\\& & 8& \color{blue}{4} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}4&2&-7&\color{orangered}{ 0 }&8&2\\& & 8& \color{orangered}{4} & & \\ \hline &2&1&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&0&8&2\\& & 8& 4& \color{blue}{16} & \\ \hline &2&1&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 16 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}4&2&-7&0&\color{orangered}{ 8 }&2\\& & 8& 4& \color{orangered}{16} & \\ \hline &2&1&4&\color{orangered}{24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 24 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-7&0&8&2\\& & 8& 4& 16& \color{blue}{96} \\ \hline &2&1&4&\color{blue}{24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 96 } = \color{orangered}{ 98 } $
$$ \begin{array}{c|rrrrr}4&2&-7&0&8&\color{orangered}{ 2 }\\& & 8& 4& 16& \color{orangered}{96} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{4}&\color{blue}{24}&\color{orangered}{98} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}+4x+24 } $ with a remainder of $ \color{red}{ 98 } $.