Divide using synthetic division $$ ( -8x^{4}+9x^{3}+40x^{2}+12x+11 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&-8&9&40&12&11\\& & -24& -45& -15& \color{black}{-9} \\ \hline &\color{blue}{-8}&\color{blue}{-15}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ -8x^{4}+9x^{3}+40x^{2}+12x+11 }{ x-3 } = \color{blue}{-8x^{3}-15x^{2}-5x-3} ~+~ \frac{ \color{red}{ 2 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&9&40&12&11\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ -8 }&9&40&12&11\\& & & & & \\ \hline &\color{orangered}{-8}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&9&40&12&11\\& & \color{blue}{-24} & & & \\ \hline &\color{blue}{-8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}3&-8&\color{orangered}{ 9 }&40&12&11\\& & \color{orangered}{-24} & & & \\ \hline &-8&\color{orangered}{-15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&9&40&12&11\\& & -24& \color{blue}{-45} & & \\ \hline &-8&\color{blue}{-15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}3&-8&9&\color{orangered}{ 40 }&12&11\\& & -24& \color{orangered}{-45} & & \\ \hline &-8&-15&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&9&40&12&11\\& & -24& -45& \color{blue}{-15} & \\ \hline &-8&-15&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&-8&9&40&\color{orangered}{ 12 }&11\\& & -24& -45& \color{orangered}{-15} & \\ \hline &-8&-15&-5&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-8&9&40&12&11\\& & -24& -45& -15& \color{blue}{-9} \\ \hline &-8&-15&-5&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&-8&9&40&12&\color{orangered}{ 11 }\\& & -24& -45& -15& \color{orangered}{-9} \\ \hline &\color{blue}{-8}&\color{blue}{-15}&\color{blue}{-5}&\color{blue}{-3}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -8x^{3}-15x^{2}-5x-3 } $ with a remainder of $ \color{red}{ 2 } $.