Divide using synthetic division $$ ( x^{4}-11x^{3}+26x^{2}+10x+28 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&1&-11&26&10&28\\& & 7& -28& -14& \color{black}{-28} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-11x^{3}+26x^{2}+10x+28 }{ x-7 } = \color{blue}{x^{3}-4x^{2}-2x-4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-11&26&10&28\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 1 }&-11&26&10&28\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-11&26&10&28\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 7 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}7&1&\color{orangered}{ -11 }&26&10&28\\& & \color{orangered}{7} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-11&26&10&28\\& & 7& \color{blue}{-28} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}7&1&-11&\color{orangered}{ 26 }&10&28\\& & 7& \color{orangered}{-28} & & \\ \hline &1&-4&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-11&26&10&28\\& & 7& -28& \color{blue}{-14} & \\ \hline &1&-4&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}7&1&-11&26&\color{orangered}{ 10 }&28\\& & 7& -28& \color{orangered}{-14} & \\ \hline &1&-4&-2&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-11&26&10&28\\& & 7& -28& -14& \color{blue}{-28} \\ \hline &1&-4&-2&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}7&1&-11&26&10&\color{orangered}{ 28 }\\& & 7& -28& -14& \color{orangered}{-28} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-4x^{2}-2x-4 } $ with a remainder of $ \color{red}{ 0 } $.