Divide using synthetic division $$ ( 4x^{3}-12x^{2}-5x-1 ) \div ( -1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&4&-12&-5&-1\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{-5}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-12x^{2}-5x-1 }{ x } = \color{blue}{4x^{2}-12x-5} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&-5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 4 }&-12&-5&-1\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&-5&-1\\& & \color{blue}{0} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}0&4&\color{orangered}{ -12 }&-5&-1\\& & \color{orangered}{0} & & \\ \hline &4&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&-5&-1\\& & 0& \color{blue}{0} & \\ \hline &4&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}0&4&-12&\color{orangered}{ -5 }&-1\\& & 0& \color{orangered}{0} & \\ \hline &4&-12&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&4&-12&-5&-1\\& & 0& 0& \color{blue}{0} \\ \hline &4&-12&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}0&4&-12&-5&\color{orangered}{ -1 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{-5}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-12x-5 } $ with a remainder of $ \color{red}{ -1 } $.