Divide using synthetic division $$ ( 7x^{3}-5x^{2}-21x+15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&7&-5&-21&15\\& & 21& 48& \color{black}{81} \\ \hline &\color{blue}{7}&\color{blue}{16}&\color{blue}{27}&\color{orangered}{96} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-5x^{2}-21x+15 }{ x-3 } = \color{blue}{7x^{2}+16x+27} ~+~ \frac{ \color{red}{ 96 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&7&-5&-21&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 7 }&-5&-21&15\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&7&-5&-21&15\\& & \color{blue}{21} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 21 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}3&7&\color{orangered}{ -5 }&-21&15\\& & \color{orangered}{21} & & \\ \hline &7&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&7&-5&-21&15\\& & 21& \color{blue}{48} & \\ \hline &7&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 48 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrr}3&7&-5&\color{orangered}{ -21 }&15\\& & 21& \color{orangered}{48} & \\ \hline &7&16&\color{orangered}{27}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&7&-5&-21&15\\& & 21& 48& \color{blue}{81} \\ \hline &7&16&\color{blue}{27}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 81 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrr}3&7&-5&-21&\color{orangered}{ 15 }\\& & 21& 48& \color{orangered}{81} \\ \hline &\color{blue}{7}&\color{blue}{16}&\color{blue}{27}&\color{orangered}{96} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+16x+27 } $ with a remainder of $ \color{red}{ 96 } $.