Divide using synthetic division $$ ( 3x^{3}-15x^{2}+7x+25 ) \div ( x-4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&-15&7&25\\& & 12& -12& \color{black}{-20} \\ \hline &\color{blue}{3}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{5} \end{array} $$The solution is:
$$ \dfrac{ 3x^{3}-15x^{2}+7x+25 }{ x-4 } = \color{blue}{3x^{2}-3x-5} ~+~ \dfrac{ \color{red}{ 5 } }{ x-4 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-15&7&25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&-15&7&25\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-15&7&25\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 12 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ -15 }&7&25\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-15&7&25\\& & 12& \color{blue}{-12} & \\ \hline &3&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&3&-15&\color{orangered}{ 7 }&25\\& & 12& \color{orangered}{-12} & \\ \hline &3&-3&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-15&7&25\\& & 12& -12& \color{blue}{-20} \\ \hline &3&-3&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}4&3&-15&7&\color{orangered}{ 25 }\\& & 12& -12& \color{orangered}{-20} \\ \hline &\color{blue}{3}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-3x-5 } $ with a remainder of $ \color{red}{ 5 } $.