Divide using synthetic division $$ ( x^{4}+11x^{2}-15x-25 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&0&11&-15&-25\\& & 3& 9& 60& \color{black}{135} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{20}&\color{blue}{45}&\color{orangered}{110} \end{array} $$The solution is:
$$ \frac{ x^{4}+11x^{2}-15x-25 }{ x-3 } = \color{blue}{x^{3}+3x^{2}+20x+45} ~+~ \frac{ \color{red}{ 110 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&11&-15&-25\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&0&11&-15&-25\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&11&-15&-25\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 0 }&11&-15&-25\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&11&-15&-25\\& & 3& \color{blue}{9} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 9 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}3&1&0&\color{orangered}{ 11 }&-15&-25\\& & 3& \color{orangered}{9} & & \\ \hline &1&3&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 20 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&11&-15&-25\\& & 3& 9& \color{blue}{60} & \\ \hline &1&3&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 60 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrrr}3&1&0&11&\color{orangered}{ -15 }&-25\\& & 3& 9& \color{orangered}{60} & \\ \hline &1&3&20&\color{orangered}{45}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 45 } = \color{blue}{ 135 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&11&-15&-25\\& & 3& 9& 60& \color{blue}{135} \\ \hline &1&3&20&\color{blue}{45}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 135 } = \color{orangered}{ 110 } $
$$ \begin{array}{c|rrrrr}3&1&0&11&-15&\color{orangered}{ -25 }\\& & 3& 9& 60& \color{orangered}{135} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{20}&\color{blue}{45}&\color{orangered}{110} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}+20x+45 } $ with a remainder of $ \color{red}{ 110 } $.