Divide using synthetic division $$ ( 3x^{4}+9x^{3}+15x^{2}-12x-20 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&3&9&15&-12&-20\\& & -15& 30& -225& \color{black}{1185} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{45}&\color{blue}{-237}&\color{orangered}{1165} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+9x^{3}+15x^{2}-12x-20 }{ x+5 } = \color{blue}{3x^{3}-6x^{2}+45x-237} ~+~ \frac{ \color{red}{ 1165 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&9&15&-12&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 3 }&9&15&-12&-20\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&9&15&-12&-20\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-5&3&\color{orangered}{ 9 }&15&-12&-20\\& & \color{orangered}{-15} & & & \\ \hline &3&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&9&15&-12&-20\\& & -15& \color{blue}{30} & & \\ \hline &3&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 30 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrrr}-5&3&9&\color{orangered}{ 15 }&-12&-20\\& & -15& \color{orangered}{30} & & \\ \hline &3&-6&\color{orangered}{45}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 45 } = \color{blue}{ -225 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&9&15&-12&-20\\& & -15& 30& \color{blue}{-225} & \\ \hline &3&-6&\color{blue}{45}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -225 \right) } = \color{orangered}{ -237 } $
$$ \begin{array}{c|rrrrr}-5&3&9&15&\color{orangered}{ -12 }&-20\\& & -15& 30& \color{orangered}{-225} & \\ \hline &3&-6&45&\color{orangered}{-237}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -237 \right) } = \color{blue}{ 1185 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&9&15&-12&-20\\& & -15& 30& -225& \color{blue}{1185} \\ \hline &3&-6&45&\color{blue}{-237}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 1185 } = \color{orangered}{ 1165 } $
$$ \begin{array}{c|rrrrr}-5&3&9&15&-12&\color{orangered}{ -20 }\\& & -15& 30& -225& \color{orangered}{1185} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{45}&\color{blue}{-237}&\color{orangered}{1165} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-6x^{2}+45x-237 } $ with a remainder of $ \color{red}{ 1165 } $.