Divide using synthetic division $$ ( 7x^{3}-32x^{2}-11x-20 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&7&-32&-11&-20\\& & 35& 15& \color{black}{20} \\ \hline &\color{blue}{7}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-32x^{2}-11x-20 }{ x-5 } = \color{blue}{7x^{2}+3x+4} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-32&-11&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 7 }&-32&-11&-20\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-32&-11&-20\\& & \color{blue}{35} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 35 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}5&7&\color{orangered}{ -32 }&-11&-20\\& & \color{orangered}{35} & & \\ \hline &7&\color{orangered}{3}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-32&-11&-20\\& & 35& \color{blue}{15} & \\ \hline &7&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 15 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}5&7&-32&\color{orangered}{ -11 }&-20\\& & 35& \color{orangered}{15} & \\ \hline &7&3&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&7&-32&-11&-20\\& & 35& 15& \color{blue}{20} \\ \hline &7&3&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&7&-32&-11&\color{orangered}{ -20 }\\& & 35& 15& \color{orangered}{20} \\ \hline &\color{blue}{7}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+3x+4 } $ with a remainder of $ \color{red}{ 0 } $.