Find remainder when $ 6x^{3}-21x^{2}-26 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&6&-21&0&-26\\& & 18& -9& \color{black}{-27} \\ \hline &\color{blue}{6}&\color{blue}{-3}&\color{blue}{-9}&\color{orangered}{-53} \end{array} $$The remainder when $ 6x^{3}-21x^{2}-26 $ is divided by $ x-3 $ is $ \, \color{red}{ -53 } $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-21&0&-26\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 6 }&-21&0&-26\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-21&0&-26\\& & \color{blue}{18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 18 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&6&\color{orangered}{ -21 }&0&-26\\& & \color{orangered}{18} & & \\ \hline &6&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-21&0&-26\\& & 18& \color{blue}{-9} & \\ \hline &6&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}3&6&-21&\color{orangered}{ 0 }&-26\\& & 18& \color{orangered}{-9} & \\ \hline &6&-3&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&-21&0&-26\\& & 18& -9& \color{blue}{-27} \\ \hline &6&-3&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrr}3&6&-21&0&\color{orangered}{ -26 }\\& & 18& -9& \color{orangered}{-27} \\ \hline &\color{blue}{6}&\color{blue}{-3}&\color{blue}{-9}&\color{orangered}{-53} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -53 }\right) $.