Divide using synthetic division $$ ( -8x^{4}+16x^{3}+28x^{2}-19x+27 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&-8&16&28&-19&27\\& & -24& -24& 12& \color{black}{-21} \\ \hline &\color{blue}{-8}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{-7}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ -8x^{4}+16x^{3}+28x^{2}-19x+27 }{ x-3 } = \color{blue}{-8x^{3}-8x^{2}+4x-7} ~+~ \frac{ \color{red}{ 6 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&16&28&-19&27\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ -8 }&16&28&-19&27\\& & & & & \\ \hline &\color{orangered}{-8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&16&28&-19&27\\& & \color{blue}{-24} & & & \\ \hline &\color{blue}{-8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&-8&\color{orangered}{ 16 }&28&-19&27\\& & \color{orangered}{-24} & & & \\ \hline &-8&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&16&28&-19&27\\& & -24& \color{blue}{-24} & & \\ \hline &-8&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&-8&16&\color{orangered}{ 28 }&-19&27\\& & -24& \color{orangered}{-24} & & \\ \hline &-8&-8&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&16&28&-19&27\\& & -24& -24& \color{blue}{12} & \\ \hline &-8&-8&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 12 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}3&-8&16&28&\color{orangered}{ -19 }&27\\& & -24& -24& \color{orangered}{12} & \\ \hline &-8&-8&4&\color{orangered}{-7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&16&28&-19&27\\& & -24& -24& 12& \color{blue}{-21} \\ \hline &-8&-8&4&\color{blue}{-7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&-8&16&28&-19&\color{orangered}{ 27 }\\& & -24& -24& 12& \color{orangered}{-21} \\ \hline &\color{blue}{-8}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{-7}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -8x^{3}-8x^{2}+4x-7 } $ with a remainder of $ \color{red}{ 6 } $.