Divide using synthetic division $$ ( 4x^{3}-2x^{2}-x+3 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&4&-2&-1&3\\& & -8& 20& \color{black}{-38} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{19}&\color{orangered}{-35} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-2x^{2}-x+3 }{ x+2 } = \color{blue}{4x^{2}-10x+19} \color{red}{~-~} \frac{ \color{red}{ 35 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-2&-1&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 4 }&-2&-1&3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-2&-1&3\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-2&4&\color{orangered}{ -2 }&-1&3\\& & \color{orangered}{-8} & & \\ \hline &4&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-2&-1&3\\& & -8& \color{blue}{20} & \\ \hline &4&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 20 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}-2&4&-2&\color{orangered}{ -1 }&3\\& & -8& \color{orangered}{20} & \\ \hline &4&-10&\color{orangered}{19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 19 } = \color{blue}{ -38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&-2&-1&3\\& & -8& 20& \color{blue}{-38} \\ \hline &4&-10&\color{blue}{19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -38 \right) } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrr}-2&4&-2&-1&\color{orangered}{ 3 }\\& & -8& 20& \color{orangered}{-38} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{19}&\color{orangered}{-35} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-10x+19 } $ with a remainder of $ \color{red}{ -35 } $.