Divide using synthetic division $$ ( 2x^{4}+x^{3}-49x^{2}+79x+15 ) \div ( 3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&2&1&-49&79&15\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-49}&\color{blue}{79}&\color{orangered}{15} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+x^{3}-49x^{2}+79x+15 }{ x } = \color{blue}{2x^{3}+x^{2}-49x+79} ~+~ \frac{ \color{red}{ 15 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&1&-49&79&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 2 }&1&-49&79&15\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&1&-49&79&15\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}0&2&\color{orangered}{ 1 }&-49&79&15\\& & \color{orangered}{0} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&1&-49&79&15\\& & 0& \color{blue}{0} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -49 } + \color{orangered}{ 0 } = \color{orangered}{ -49 } $
$$ \begin{array}{c|rrrrr}0&2&1&\color{orangered}{ -49 }&79&15\\& & 0& \color{orangered}{0} & & \\ \hline &2&1&\color{orangered}{-49}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -49 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&1&-49&79&15\\& & 0& 0& \color{blue}{0} & \\ \hline &2&1&\color{blue}{-49}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 79 } + \color{orangered}{ 0 } = \color{orangered}{ 79 } $
$$ \begin{array}{c|rrrrr}0&2&1&-49&\color{orangered}{ 79 }&15\\& & 0& 0& \color{orangered}{0} & \\ \hline &2&1&-49&\color{orangered}{79}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 79 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&2&1&-49&79&15\\& & 0& 0& 0& \color{blue}{0} \\ \hline &2&1&-49&\color{blue}{79}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 0 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}0&2&1&-49&79&\color{orangered}{ 15 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-49}&\color{blue}{79}&\color{orangered}{15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}-49x+79 } $ with a remainder of $ \color{red}{ 15 } $.