Find remainder when $ 3x^{3}+18x^{2}+19x+13 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&3&18&19&13\\& & -15& -15& \color{black}{-20} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{-7} \end{array} $$The remainder when $ 3x^{3}+18x^{2}+19x+13 $ is divided by $ x+5 $ is $ \, \color{red}{ -7 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&18&19&13\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 3 }&18&19&13\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&18&19&13\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-5&3&\color{orangered}{ 18 }&19&13\\& & \color{orangered}{-15} & & \\ \hline &3&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&18&19&13\\& & -15& \color{blue}{-15} & \\ \hline &3&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-5&3&18&\color{orangered}{ 19 }&13\\& & -15& \color{orangered}{-15} & \\ \hline &3&3&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&3&18&19&13\\& & -15& -15& \color{blue}{-20} \\ \hline &3&3&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-5&3&18&19&\color{orangered}{ 13 }\\& & -15& -15& \color{orangered}{-20} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{4}&\color{orangered}{-7} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -7 }\right) $.