Divide using synthetic division $$ ( -4x^{3}-6x+3 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&-4&0&-6&3\\& & -8& -16& \color{black}{-44} \\ \hline &\color{blue}{-4}&\color{blue}{-8}&\color{blue}{-22}&\color{orangered}{-41} \end{array} $$The solution is:
$$ \frac{ -4x^{3}-6x+3 }{ x-2 } = \color{blue}{-4x^{2}-8x-22} \color{red}{~-~} \frac{ \color{red}{ 41 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&0&-6&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ -4 }&0&-6&3\\& & & & \\ \hline &\color{orangered}{-4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&0&-6&3\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{-4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}2&-4&\color{orangered}{ 0 }&-6&3\\& & \color{orangered}{-8} & & \\ \hline &-4&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&0&-6&3\\& & -8& \color{blue}{-16} & \\ \hline &-4&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrr}2&-4&0&\color{orangered}{ -6 }&3\\& & -8& \color{orangered}{-16} & \\ \hline &-4&-8&\color{orangered}{-22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -44 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&-4&0&-6&3\\& & -8& -16& \color{blue}{-44} \\ \hline &-4&-8&\color{blue}{-22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -44 \right) } = \color{orangered}{ -41 } $
$$ \begin{array}{c|rrrr}2&-4&0&-6&\color{orangered}{ 3 }\\& & -8& -16& \color{orangered}{-44} \\ \hline &\color{blue}{-4}&\color{blue}{-8}&\color{blue}{-22}&\color{orangered}{-41} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{2}-8x-22 } $ with a remainder of $ \color{red}{ -41 } $.