Find remainder when $ -7x^{3}+22x^{2}-9 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-7&22&0&-9\\& & -21& 3& \color{black}{9} \\ \hline &\color{blue}{-7}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The remainder when $ -7x^{3}+22x^{2}-9 $ is divided by $ x-3 $ is $ \, \color{red}{ 0 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&22&0&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -7 }&22&0&-9\\& & & & \\ \hline &\color{orangered}{-7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&22&0&-9\\& & \color{blue}{-21} & & \\ \hline &\color{blue}{-7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&-7&\color{orangered}{ 22 }&0&-9\\& & \color{orangered}{-21} & & \\ \hline &-7&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&22&0&-9\\& & -21& \color{blue}{3} & \\ \hline &-7&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&-7&22&\color{orangered}{ 0 }&-9\\& & -21& \color{orangered}{3} & \\ \hline &-7&1&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-7&22&0&-9\\& & -21& 3& \color{blue}{9} \\ \hline &-7&1&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 9 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&-7&22&0&\color{orangered}{ -9 }\\& & -21& 3& \color{orangered}{9} \\ \hline &\color{blue}{-7}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.