Divide using synthetic division $$ ( 2x^{3}+x^{2}-7x-240 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&1&-7&-240\\& & 10& 55& \color{black}{240} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{48}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+x^{2}-7x-240 }{ x-5 } = \color{blue}{2x^{2}+11x+48} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&1&-7&-240\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&1&-7&-240\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&1&-7&-240\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 10 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ 1 }&-7&-240\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 11 } = \color{blue}{ 55 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&1&-7&-240\\& & 10& \color{blue}{55} & \\ \hline &2&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 55 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}5&2&1&\color{orangered}{ -7 }&-240\\& & 10& \color{orangered}{55} & \\ \hline &2&11&\color{orangered}{48}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 48 } = \color{blue}{ 240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&1&-7&-240\\& & 10& 55& \color{blue}{240} \\ \hline &2&11&\color{blue}{48}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -240 } + \color{orangered}{ 240 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&2&1&-7&\color{orangered}{ -240 }\\& & 10& 55& \color{orangered}{240} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{48}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+11x+48 } $ with a remainder of $ \color{red}{ 0 } $.