Find remainder when $ 5x^{2}-4x+9 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&5&-4&9\\& & 15& \color{black}{33} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{orangered}{42} \end{array} $$The remainder when $ 5x^{2}-4x+9 $ is divided by $ x-3 $ is $ \, \color{red}{ 42 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrr}\color{blue}{3}&5&-4&9\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 5 }&-4&9\\& & & \\ \hline &\color{orangered}{5}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&5&-4&9\\& & \color{blue}{15} & \\ \hline &\color{blue}{5}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 15 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrr}3&5&\color{orangered}{ -4 }&9\\& & \color{orangered}{15} & \\ \hline &5&\color{orangered}{11}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 11 } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&5&-4&9\\& & 15& \color{blue}{33} \\ \hline &5&\color{blue}{11}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 33 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrr}3&5&-4&\color{orangered}{ 9 }\\& & 15& \color{orangered}{33} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{orangered}{42} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 42 }\right) $.