Divide using synthetic division $$ ( 2x^{3}-x^{2}-16x-9 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&-1&-16&-9\\& & -6& 21& \color{black}{-15} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{-24} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-x^{2}-16x-9 }{ x+3 } = \color{blue}{2x^{2}-7x+5} \color{red}{~-~} \frac{ \color{red}{ 24 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&-16&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&-1&-16&-9\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&-16&-9\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ -1 }&-16&-9\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&-16&-9\\& & -6& \color{blue}{21} & \\ \hline &2&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 21 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-3&2&-1&\color{orangered}{ -16 }&-9\\& & -6& \color{orangered}{21} & \\ \hline &2&-7&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&-16&-9\\& & -6& 21& \color{blue}{-15} \\ \hline &2&-7&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-3&2&-1&-16&\color{orangered}{ -9 }\\& & -6& 21& \color{orangered}{-15} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{-24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-7x+5 } $ with a remainder of $ \color{red}{ -24 } $.