Divide using synthetic division $$ ( 2x^{4}+13x^{3}+14x^{2}-30x-27 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&13&14&-30&-27\\& & -6& -21& 21& \color{black}{27} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+13x^{3}+14x^{2}-30x-27 }{ x+3 } = \color{blue}{2x^{3}+7x^{2}-7x-9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&13&14&-30&-27\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&13&14&-30&-27\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&13&14&-30&-27\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ 13 }&14&-30&-27\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&13&14&-30&-27\\& & -6& \color{blue}{-21} & & \\ \hline &2&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-3&2&13&\color{orangered}{ 14 }&-30&-27\\& & -6& \color{orangered}{-21} & & \\ \hline &2&7&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&13&14&-30&-27\\& & -6& -21& \color{blue}{21} & \\ \hline &2&7&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 21 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-3&2&13&14&\color{orangered}{ -30 }&-27\\& & -6& -21& \color{orangered}{21} & \\ \hline &2&7&-7&\color{orangered}{-9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&13&14&-30&-27\\& & -6& -21& 21& \color{blue}{27} \\ \hline &2&7&-7&\color{blue}{-9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 27 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&2&13&14&-30&\color{orangered}{ -27 }\\& & -6& -21& 21& \color{orangered}{27} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+7x^{2}-7x-9 } $ with a remainder of $ \color{red}{ 0 } $.