Divide using synthetic division $$ ( 2x^{4}+8x^{3}-13x^{2}-16x-5 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&2&8&-13&-16&-5\\& & -10& 10& 15& \color{black}{5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-3}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+8x^{3}-13x^{2}-16x-5 }{ x+5 } = \color{blue}{2x^{3}-2x^{2}-3x-1} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&8&-13&-16&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 2 }&8&-13&-16&-5\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&8&-13&-16&-5\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-5&2&\color{orangered}{ 8 }&-13&-16&-5\\& & \color{orangered}{-10} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&8&-13&-16&-5\\& & -10& \color{blue}{10} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 10 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-5&2&8&\color{orangered}{ -13 }&-16&-5\\& & -10& \color{orangered}{10} & & \\ \hline &2&-2&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&8&-13&-16&-5\\& & -10& 10& \color{blue}{15} & \\ \hline &2&-2&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 15 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-5&2&8&-13&\color{orangered}{ -16 }&-5\\& & -10& 10& \color{orangered}{15} & \\ \hline &2&-2&-3&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&8&-13&-16&-5\\& & -10& 10& 15& \color{blue}{5} \\ \hline &2&-2&-3&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 5 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&2&8&-13&-16&\color{orangered}{ -5 }\\& & -10& 10& 15& \color{orangered}{5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-3}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x^{2}-3x-1 } $ with a remainder of $ \color{red}{ 0 } $.