Divide using synthetic division $$ ( 5x^{4}+11x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&5&0&0&11&0\\& & 15& 45& 135& \color{black}{438} \\ \hline &\color{blue}{5}&\color{blue}{15}&\color{blue}{45}&\color{blue}{146}&\color{orangered}{438} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+11x }{ x-3 } = \color{blue}{5x^{3}+15x^{2}+45x+146} ~+~ \frac{ \color{red}{ 438 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&0&0&11&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 5 }&0&0&11&0\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&0&0&11&0\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}3&5&\color{orangered}{ 0 }&0&11&0\\& & \color{orangered}{15} & & & \\ \hline &5&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&0&0&11&0\\& & 15& \color{blue}{45} & & \\ \hline &5&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 45 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrrr}3&5&0&\color{orangered}{ 0 }&11&0\\& & 15& \color{orangered}{45} & & \\ \hline &5&15&\color{orangered}{45}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 45 } = \color{blue}{ 135 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&0&0&11&0\\& & 15& 45& \color{blue}{135} & \\ \hline &5&15&\color{blue}{45}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 135 } = \color{orangered}{ 146 } $
$$ \begin{array}{c|rrrrr}3&5&0&0&\color{orangered}{ 11 }&0\\& & 15& 45& \color{orangered}{135} & \\ \hline &5&15&45&\color{orangered}{146}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 146 } = \color{blue}{ 438 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&0&0&11&0\\& & 15& 45& 135& \color{blue}{438} \\ \hline &5&15&45&\color{blue}{146}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 438 } = \color{orangered}{ 438 } $
$$ \begin{array}{c|rrrrr}3&5&0&0&11&\color{orangered}{ 0 }\\& & 15& 45& 135& \color{orangered}{438} \\ \hline &\color{blue}{5}&\color{blue}{15}&\color{blue}{45}&\color{blue}{146}&\color{orangered}{438} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+15x^{2}+45x+146 } $ with a remainder of $ \color{red}{ 438 } $.