Find remainder when $ 2x^{3}+x^{2}-23x+20 $ is divided by $ x+4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&2&1&-23&20\\& & -8& 28& \color{black}{-20} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The remainder when $ 2x^{3}+x^{2}-23x+20 $ is divided by $ x+4 $ is $ \, \color{red}{ 0 } $.
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 + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&1&-23&20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 2 }&1&-23&20\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&1&-23&20\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-4&2&\color{orangered}{ 1 }&-23&20\\& & \color{orangered}{-8} & & \\ \hline &2&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&1&-23&20\\& & -8& \color{blue}{28} & \\ \hline &2&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 28 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-4&2&1&\color{orangered}{ -23 }&20\\& & -8& \color{orangered}{28} & \\ \hline &2&-7&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&2&1&-23&20\\& & -8& 28& \color{blue}{-20} \\ \hline &2&-7&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&2&1&-23&\color{orangered}{ 20 }\\& & -8& 28& \color{orangered}{-20} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.