Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&7&-5&-21&15\\& & 49& 308& \color{black}{2009} \\ \hline &\color{blue}{7}&\color{blue}{44}&\color{blue}{287}&\color{orangered}{2024} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x-7 } = \color{blue}{7x^{2}+44x+287} ~+~ \frac{ \color{red}{ 2024 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 7 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&7&-5&-21&15\\& & \color{blue}{49} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 49 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrr}7&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{49} & & \\ \hline &7&\color{orangered}{44}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 44 } = \color{blue}{ 308 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&7&-5&-21&15\\& & 49& \color{blue}{308} & \\ \hline &7&\color{blue}{44}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 308 } = \color{orangered}{ 287 } $
$$ \begin{array}{c|rrrr}7&7&-5&\color{orangered}{ -21 }&15\\& & 49& \color{orangered}{308} & \\ \hline &7&44&\color{orangered}{287}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 287 } = \color{blue}{ 2009 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&7&-5&-21&15\\& & 49& 308& \color{blue}{2009} \\ \hline &7&44&\color{blue}{287}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 2009 } = \color{orangered}{ 2024 } $
$$ \begin{array}{c|rrrr}7&7&-5&-21&\color{orangered}{ 15 }\\& & 49& 308& \color{orangered}{2009} \\ \hline &\color{blue}{7}&\color{blue}{44}&\color{blue}{287}&\color{orangered}{2024} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+44x+287 } $ with a remainder of $ \color{red}{ 2024 } $.