Divide using synthetic division $$ ( 2x^{3}+5x^{2}-18x-45 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&5&-18&-45\\& & 10& 75& \color{black}{285} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{57}&\color{orangered}{240} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+5x^{2}-18x-45 }{ x-5 } = \color{blue}{2x^{2}+15x+57} ~+~ \frac{ \color{red}{ 240 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&5&-18&-45\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&5&-18&-45\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&5&-18&-45\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 10 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ 5 }&-18&-45\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 15 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&5&-18&-45\\& & 10& \color{blue}{75} & \\ \hline &2&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 75 } = \color{orangered}{ 57 } $
$$ \begin{array}{c|rrrr}5&2&5&\color{orangered}{ -18 }&-45\\& & 10& \color{orangered}{75} & \\ \hline &2&15&\color{orangered}{57}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 57 } = \color{blue}{ 285 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&5&-18&-45\\& & 10& 75& \color{blue}{285} \\ \hline &2&15&\color{blue}{57}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 285 } = \color{orangered}{ 240 } $
$$ \begin{array}{c|rrrr}5&2&5&-18&\color{orangered}{ -45 }\\& & 10& 75& \color{orangered}{285} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{57}&\color{orangered}{240} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+15x+57 } $ with a remainder of $ \color{red}{ 240 } $.