Divide using synthetic division $$ ( 2x^{3}-2x-3 ) \div ( 2x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&0&-2&-3\\& & -10& 50& \color{black}{-240} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{48}&\color{orangered}{-243} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-2x-3 }{ x+5 } = \color{blue}{2x^{2}-10x+48} \color{red}{~-~} \frac{ \color{red}{ 243 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&-2&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&0&-2&-3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&-2&-3\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 0 }&-2&-3\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&-2&-3\\& & -10& \color{blue}{50} & \\ \hline &2&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 50 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}-5&2&0&\color{orangered}{ -2 }&-3\\& & -10& \color{orangered}{50} & \\ \hline &2&-10&\color{orangered}{48}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 48 } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&-2&-3\\& & -10& 50& \color{blue}{-240} \\ \hline &2&-10&\color{blue}{48}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ -243 } $
$$ \begin{array}{c|rrrr}-5&2&0&-2&\color{orangered}{ -3 }\\& & -10& 50& \color{orangered}{-240} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{48}&\color{orangered}{-243} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-10x+48 } $ with a remainder of $ \color{red}{ -243 } $.