Divide using synthetic division $$ ( 27x^{3}-3x+5 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&27&0&-3&5\\& & -27& 27& \color{black}{-24} \\ \hline &\color{blue}{27}&\color{blue}{-27}&\color{blue}{24}&\color{orangered}{-19} \end{array} $$The solution is:
$$ \frac{ 27x^{3}-3x+5 }{ x+1 } = \color{blue}{27x^{2}-27x+24} \color{red}{~-~} \frac{ \color{red}{ 19 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&27&0&-3&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 27 }&0&-3&5\\& & & & \\ \hline &\color{orangered}{27}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 27 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&27&0&-3&5\\& & \color{blue}{-27} & & \\ \hline &\color{blue}{27}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrr}-1&27&\color{orangered}{ 0 }&-3&5\\& & \color{orangered}{-27} & & \\ \hline &27&\color{orangered}{-27}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&27&0&-3&5\\& & -27& \color{blue}{27} & \\ \hline &27&\color{blue}{-27}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 27 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}-1&27&0&\color{orangered}{ -3 }&5\\& & -27& \color{orangered}{27} & \\ \hline &27&-27&\color{orangered}{24}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 24 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&27&0&-3&5\\& & -27& 27& \color{blue}{-24} \\ \hline &27&-27&\color{blue}{24}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-1&27&0&-3&\color{orangered}{ 5 }\\& & -27& 27& \color{orangered}{-24} \\ \hline &\color{blue}{27}&\color{blue}{-27}&\color{blue}{24}&\color{orangered}{-19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 27x^{2}-27x+24 } $ with a remainder of $ \color{red}{ -19 } $.