Divide using synthetic division $$ ( 4x^{3}-18x-5 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&4&0&-18&-5\\& & -8& 16& \color{black}{4} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{-2}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-18x-5 }{ x+2 } = \color{blue}{4x^{2}-8x-2} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&0&-18&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 4 }&0&-18&-5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&0&-18&-5\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-2&4&\color{orangered}{ 0 }&-18&-5\\& & \color{orangered}{-8} & & \\ \hline &4&\color{orangered}{-8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&0&-18&-5\\& & -8& \color{blue}{16} & \\ \hline &4&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 16 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-2&4&0&\color{orangered}{ -18 }&-5\\& & -8& \color{orangered}{16} & \\ \hline &4&-8&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&4&0&-18&-5\\& & -8& 16& \color{blue}{4} \\ \hline &4&-8&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 4 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-2&4&0&-18&\color{orangered}{ -5 }\\& & -8& 16& \color{orangered}{4} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{-2}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-8x-2 } $ with a remainder of $ \color{red}{ -1 } $.