Divide using synthetic division $$ ( -21x^{2}-40x+48 ) \div ( -4x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}0&-21&-40&48\\& & 0& \color{black}{0} \\ \hline &\color{blue}{-21}&\color{blue}{-40}&\color{orangered}{48} \end{array} $$The solution is:
$$ \frac{ -21x^{2}-40x+48 }{ x } = \color{blue}{-21x-40} ~+~ \frac{ \color{red}{ 48 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrr}\color{blue}{0}&-21&-40&48\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}0&\color{orangered}{ -21 }&-40&48\\& & & \\ \hline &\color{orangered}{-21}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&-21&-40&48\\& & \color{blue}{0} & \\ \hline &\color{blue}{-21}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 0 } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrr}0&-21&\color{orangered}{ -40 }&48\\& & \color{orangered}{0} & \\ \hline &-21&\color{orangered}{-40}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&-21&-40&48\\& & 0& \color{blue}{0} \\ \hline &-21&\color{blue}{-40}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ 0 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrr}0&-21&-40&\color{orangered}{ 48 }\\& & 0& \color{orangered}{0} \\ \hline &\color{blue}{-21}&\color{blue}{-40}&\color{orangered}{48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -21x-40 } $ with a remainder of $ \color{red}{ 48 } $.