Divide using synthetic division $$ ( x^{5}+2x^{4}-22x^{3}+24x^{2}+x-30 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&2&-22&24&1&-30\\& & 3& 15& -21& 9& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-7}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+2x^{4}-22x^{3}+24x^{2}+x-30 }{ x-3 } = \color{blue}{x^{4}+5x^{3}-7x^{2}+3x+10} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&2&-22&24&1&-30\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 2 }&-22&24&1&-30\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & 3& \color{blue}{15} & & & \\ \hline &1&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 15 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}3&1&2&\color{orangered}{ -22 }&24&1&-30\\& & 3& \color{orangered}{15} & & & \\ \hline &1&5&\color{orangered}{-7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & 3& 15& \color{blue}{-21} & & \\ \hline &1&5&\color{blue}{-7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-22&\color{orangered}{ 24 }&1&-30\\& & 3& 15& \color{orangered}{-21} & & \\ \hline &1&5&-7&\color{orangered}{3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & 3& 15& -21& \color{blue}{9} & \\ \hline &1&5&-7&\color{blue}{3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-22&24&\color{orangered}{ 1 }&-30\\& & 3& 15& -21& \color{orangered}{9} & \\ \hline &1&5&-7&3&\color{orangered}{10}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-22&24&1&-30\\& & 3& 15& -21& 9& \color{blue}{30} \\ \hline &1&5&-7&3&\color{blue}{10}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 30 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-22&24&1&\color{orangered}{ -30 }\\& & 3& 15& -21& 9& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-7}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+5x^{3}-7x^{2}+3x+10 } $ with a remainder of $ \color{red}{ 0 } $.