Divide using synthetic division $$ ( -4x^{3}+6x^{2}-7 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-4&6&0&-7\\& & -12& -18& \color{black}{-54} \\ \hline &\color{blue}{-4}&\color{blue}{-6}&\color{blue}{-18}&\color{orangered}{-61} \end{array} $$The solution is:
$$ \frac{ -4x^{3}+6x^{2}-7 }{ x-3 } = \color{blue}{-4x^{2}-6x-18} \color{red}{~-~} \frac{ \color{red}{ 61 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&6&0&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -4 }&6&0&-7\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&6&0&-7\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}3&-4&\color{orangered}{ 6 }&0&-7\\& & \color{orangered}{-12} & & \\ \hline &-4&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&6&0&-7\\& & -12& \color{blue}{-18} & \\ \hline &-4&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}3&-4&6&\color{orangered}{ 0 }&-7\\& & -12& \color{orangered}{-18} & \\ \hline &-4&-6&\color{orangered}{-18}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-4&6&0&-7\\& & -12& -18& \color{blue}{-54} \\ \hline &-4&-6&\color{blue}{-18}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -61 } $
$$ \begin{array}{c|rrrr}3&-4&6&0&\color{orangered}{ -7 }\\& & -12& -18& \color{orangered}{-54} \\ \hline &\color{blue}{-4}&\color{blue}{-6}&\color{blue}{-18}&\color{orangered}{-61} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{2}-6x-18 } $ with a remainder of $ \color{red}{ -61 } $.