Find remainder when $ 2x^{3}+38x+22 $ is divided by $ 2x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&0&38&22\\& & -10& 50& \color{black}{-440} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{88}&\color{orangered}{-418} \end{array} $$The remainder when $ 2x^{3}+38x+22 $ is divided by $ x+5 $ is $ \, \color{red}{ -418 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&38&22\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&0&38&22\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&38&22\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 0 }&38&22\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{-10}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&38&22\\& & -10& \color{blue}{50} & \\ \hline &2&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 38 } + \color{orangered}{ 50 } = \color{orangered}{ 88 } $
$$ \begin{array}{c|rrrr}-5&2&0&\color{orangered}{ 38 }&22\\& & -10& \color{orangered}{50} & \\ \hline &2&-10&\color{orangered}{88}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 88 } = \color{blue}{ -440 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&0&38&22\\& & -10& 50& \color{blue}{-440} \\ \hline &2&-10&\color{blue}{88}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -440 \right) } = \color{orangered}{ -418 } $
$$ \begin{array}{c|rrrr}-5&2&0&38&\color{orangered}{ 22 }\\& & -10& 50& \color{orangered}{-440} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{88}&\color{orangered}{-418} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -418 }\right) $.