Divide using synthetic division $$ ( x^{3}-13x^{2}+40x ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&-13&40&0\\& & 7& -42& \color{black}{-14} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-2}&\color{orangered}{-14} \end{array} $$The solution is:
$$ \frac{ x^{3}-13x^{2}+40x }{ x-7 } = \color{blue}{x^{2}-6x-2} \color{red}{~-~} \frac{ \color{red}{ 14 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-13&40&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&-13&40&0\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-13&40&0\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 7 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ -13 }&40&0\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-13&40&0\\& & 7& \color{blue}{-42} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}7&1&-13&\color{orangered}{ 40 }&0\\& & 7& \color{orangered}{-42} & \\ \hline &1&-6&\color{orangered}{-2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-13&40&0\\& & 7& -42& \color{blue}{-14} \\ \hline &1&-6&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}7&1&-13&40&\color{orangered}{ 0 }\\& & 7& -42& \color{orangered}{-14} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-2}&\color{orangered}{-14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x-2 } $ with a remainder of $ \color{red}{ -14 } $.