Divide using synthetic division $$ ( 5x^{3}+9x^{2}-4x-5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&5&9&-4&-5\\& & -15& 18& \color{black}{-42} \\ \hline &\color{blue}{5}&\color{blue}{-6}&\color{blue}{14}&\color{orangered}{-47} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+9x^{2}-4x-5 }{ x+3 } = \color{blue}{5x^{2}-6x+14} \color{red}{~-~} \frac{ \color{red}{ 47 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&9&-4&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 5 }&9&-4&-5\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&9&-4&-5\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&5&\color{orangered}{ 9 }&-4&-5\\& & \color{orangered}{-15} & & \\ \hline &5&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&9&-4&-5\\& & -15& \color{blue}{18} & \\ \hline &5&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 18 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-3&5&9&\color{orangered}{ -4 }&-5\\& & -15& \color{orangered}{18} & \\ \hline &5&-6&\color{orangered}{14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 14 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&9&-4&-5\\& & -15& 18& \color{blue}{-42} \\ \hline &5&-6&\color{blue}{14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrr}-3&5&9&-4&\color{orangered}{ -5 }\\& & -15& 18& \color{orangered}{-42} \\ \hline &\color{blue}{5}&\color{blue}{-6}&\color{blue}{14}&\color{orangered}{-47} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-6x+14 } $ with a remainder of $ \color{red}{ -47 } $.