Divide using synthetic division $$ ( 2x^{3}+8x^{2}-3x-10 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&2&8&-3&-10\\& & -8& 0& \color{black}{12} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+8x^{2}-3x-10 }{ x+4 } = \color{blue}{2x^{2}-3} ~+~ \frac{ \color{red}{ 2 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&8&-3&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 2 }&8&-3&-10\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&8&-3&-10\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&2&\color{orangered}{ 8 }&-3&-10\\& & \color{orangered}{-8} & & \\ \hline &2&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&8&-3&-10\\& & -8& \color{blue}{0} & \\ \hline &2&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-4&2&8&\color{orangered}{ -3 }&-10\\& & -8& \color{orangered}{0} & \\ \hline &2&0&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&8&-3&-10\\& & -8& 0& \color{blue}{12} \\ \hline &2&0&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-4&2&8&-3&\color{orangered}{ -10 }\\& & -8& 0& \color{orangered}{12} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-3 } $ with a remainder of $ \color{red}{ 2 } $.