Divide using synthetic division $$ ( -4x^{3}+11x^{2}+8x-10 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-4&11&8&-10\\& & -12& -3& \color{black}{15} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ -4x^{3}+11x^{2}+8x-10 }{ x-3 } = \color{blue}{-4x^{2}-x+5} ~+~ \frac{ \color{red}{ 5 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&11&8&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -4 }&11&8&-10\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&11&8&-10\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}3&-4&\color{orangered}{ 11 }&8&-10\\& & \color{orangered}{-12} & & \\ \hline &-4&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&11&8&-10\\& & -12& \color{blue}{-3} & \\ \hline &-4&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&-4&11&\color{orangered}{ 8 }&-10\\& & -12& \color{orangered}{-3} & \\ \hline &-4&-1&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&11&8&-10\\& & -12& -3& \color{blue}{15} \\ \hline &-4&-1&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 15 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&-4&11&8&\color{orangered}{ -10 }\\& & -12& -3& \color{orangered}{15} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{2}-x+5 } $ with a remainder of $ \color{red}{ 5 } $.