Divide using synthetic division $$ ( 10x^{3}+27x^{2}-6x+9 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&10&27&-6&9\\& & -30& 9& \color{black}{-9} \\ \hline &\color{blue}{10}&\color{blue}{-3}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 10x^{3}+27x^{2}-6x+9 }{ x+3 } = \color{blue}{10x^{2}-3x+3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&10&27&-6&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 10 }&27&-6&9\\& & & & \\ \hline &\color{orangered}{10}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&10&27&-6&9\\& & \color{blue}{-30} & & \\ \hline &\color{blue}{10}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-3&10&\color{orangered}{ 27 }&-6&9\\& & \color{orangered}{-30} & & \\ \hline &10&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&10&27&-6&9\\& & -30& \color{blue}{9} & \\ \hline &10&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 9 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-3&10&27&\color{orangered}{ -6 }&9\\& & -30& \color{orangered}{9} & \\ \hline &10&-3&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&10&27&-6&9\\& & -30& 9& \color{blue}{-9} \\ \hline &10&-3&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-3&10&27&-6&\color{orangered}{ 9 }\\& & -30& 9& \color{orangered}{-9} \\ \hline &\color{blue}{10}&\color{blue}{-3}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{2}-3x+3 } $ with a remainder of $ \color{red}{ 0 } $.