Divide using synthetic division $$ ( x^{4}+11x^{3}+26x^{2}-21x-5 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&11&26&-21&-5\\& & 5& 80& 530& \color{black}{2545} \\ \hline &\color{blue}{1}&\color{blue}{16}&\color{blue}{106}&\color{blue}{509}&\color{orangered}{2540} \end{array} $$The solution is:
$$ \frac{ x^{4}+11x^{3}+26x^{2}-21x-5 }{ x-5 } = \color{blue}{x^{3}+16x^{2}+106x+509} ~+~ \frac{ \color{red}{ 2540 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&11&26&-21&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&11&26&-21&-5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&11&26&-21&-5\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 5 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 11 }&26&-21&-5\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 16 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&11&26&-21&-5\\& & 5& \color{blue}{80} & & \\ \hline &1&\color{blue}{16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ 80 } = \color{orangered}{ 106 } $
$$ \begin{array}{c|rrrrr}5&1&11&\color{orangered}{ 26 }&-21&-5\\& & 5& \color{orangered}{80} & & \\ \hline &1&16&\color{orangered}{106}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 106 } = \color{blue}{ 530 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&11&26&-21&-5\\& & 5& 80& \color{blue}{530} & \\ \hline &1&16&\color{blue}{106}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 530 } = \color{orangered}{ 509 } $
$$ \begin{array}{c|rrrrr}5&1&11&26&\color{orangered}{ -21 }&-5\\& & 5& 80& \color{orangered}{530} & \\ \hline &1&16&106&\color{orangered}{509}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 509 } = \color{blue}{ 2545 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&11&26&-21&-5\\& & 5& 80& 530& \color{blue}{2545} \\ \hline &1&16&106&\color{blue}{509}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 2545 } = \color{orangered}{ 2540 } $
$$ \begin{array}{c|rrrrr}5&1&11&26&-21&\color{orangered}{ -5 }\\& & 5& 80& 530& \color{orangered}{2545} \\ \hline &\color{blue}{1}&\color{blue}{16}&\color{blue}{106}&\color{blue}{509}&\color{orangered}{2540} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+16x^{2}+106x+509 } $ with a remainder of $ \color{red}{ 2540 } $.