Divide using synthetic division $$ ( 9x^{3}+45x^{2}-5x-25 ) \div ( x+5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&9&45&-5&-25\\& & -45& 0& \color{black}{25} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 9x^{3}+45x^{2}-5x-25 }{ x+5 } = \color{blue}{9x^{2}-5} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&9&45&-5&-25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 9 }&45&-5&-25\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 9 } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&9&45&-5&-25\\& & \color{blue}{-45} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-5&9&\color{orangered}{ 45 }&-5&-25\\& & \color{orangered}{-45} & & \\ \hline &9&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&9&45&-5&-25\\& & -45& \color{blue}{0} & \\ \hline &9&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-5&9&45&\color{orangered}{ -5 }&-25\\& & -45& \color{orangered}{0} & \\ \hline &9&0&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&9&45&-5&-25\\& & -45& 0& \color{blue}{25} \\ \hline &9&0&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 25 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-5&9&45&-5&\color{orangered}{ -25 }\\& & -45& 0& \color{orangered}{25} \\ \hline &\color{blue}{9}&\color{blue}{0}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-5 } $ with a remainder of $ \color{red}{ 0 } $.