Divide using synthetic division $$ ( -6x^{3}+x^{2}-4 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-6&1&0&-4\\& & -18& -51& \color{black}{-153} \\ \hline &\color{blue}{-6}&\color{blue}{-17}&\color{blue}{-51}&\color{orangered}{-157} \end{array} $$The solution is:
$$ \frac{ -6x^{3}+x^{2}-4 }{ x-3 } = \color{blue}{-6x^{2}-17x-51} \color{red}{~-~} \frac{ \color{red}{ 157 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-6&1&0&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -6 }&1&0&-4\\& & & & \\ \hline &\color{orangered}{-6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-6&1&0&-4\\& & \color{blue}{-18} & & \\ \hline &\color{blue}{-6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}3&-6&\color{orangered}{ 1 }&0&-4\\& & \color{orangered}{-18} & & \\ \hline &-6&\color{orangered}{-17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ -51 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-6&1&0&-4\\& & -18& \color{blue}{-51} & \\ \hline &-6&\color{blue}{-17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -51 \right) } = \color{orangered}{ -51 } $
$$ \begin{array}{c|rrrr}3&-6&1&\color{orangered}{ 0 }&-4\\& & -18& \color{orangered}{-51} & \\ \hline &-6&-17&\color{orangered}{-51}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -51 \right) } = \color{blue}{ -153 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-6&1&0&-4\\& & -18& -51& \color{blue}{-153} \\ \hline &-6&-17&\color{blue}{-51}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -153 \right) } = \color{orangered}{ -157 } $
$$ \begin{array}{c|rrrr}3&-6&1&0&\color{orangered}{ -4 }\\& & -18& -51& \color{orangered}{-153} \\ \hline &\color{blue}{-6}&\color{blue}{-17}&\color{blue}{-51}&\color{orangered}{-157} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{2}-17x-51 } $ with a remainder of $ \color{red}{ -157 } $.