Divide using synthetic division $$ ( x^{4}-2x^{3}-22x^{2}+40x-21 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-2&-22&40&-21\\& & 5& 15& -35& \color{black}{25} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-22x^{2}+40x-21 }{ x-5 } = \color{blue}{x^{3}+3x^{2}-7x+5} ~+~ \frac{ \color{red}{ 4 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&-22&40&-21\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-2&-22&40&-21\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&-22&40&-21\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 5 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -2 }&-22&40&-21\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&-22&40&-21\\& & 5& \color{blue}{15} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 15 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}5&1&-2&\color{orangered}{ -22 }&40&-21\\& & 5& \color{orangered}{15} & & \\ \hline &1&3&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&-22&40&-21\\& & 5& 15& \color{blue}{-35} & \\ \hline &1&3&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&1&-2&-22&\color{orangered}{ 40 }&-21\\& & 5& 15& \color{orangered}{-35} & \\ \hline &1&3&-7&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-2&-22&40&-21\\& & 5& 15& -35& \color{blue}{25} \\ \hline &1&3&-7&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 25 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&1&-2&-22&40&\color{orangered}{ -21 }\\& & 5& 15& -35& \color{orangered}{25} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-7x+5 } $ with a remainder of $ \color{red}{ 4 } $.