Divide using synthetic division $$ ( 2x^{3}+5x^{2}-18x-45 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&5&-18&-45\\& & -10& 25& \color{black}{-35} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{7}&\color{orangered}{-80} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+5x^{2}-18x-45 }{ x+5 } = \color{blue}{2x^{2}-5x+7} \color{red}{~-~} \frac{ \color{red}{ 80 } }{ x+5 } $$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}{ \left( -10 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 5 }&-18&-45\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&5&-18&-45\\& & -10& \color{blue}{25} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 25 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-5&2&5&\color{orangered}{ -18 }&-45\\& & -10& \color{orangered}{25} & \\ \hline &2&-5&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 7 } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&5&-18&-45\\& & -10& 25& \color{blue}{-35} \\ \hline &2&-5&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrr}-5&2&5&-18&\color{orangered}{ -45 }\\& & -10& 25& \color{orangered}{-35} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{7}&\color{orangered}{-80} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-5x+7 } $ with a remainder of $ \color{red}{ -80 } $.