Divide using synthetic division $$ ( 2x^{4}-11x^{3}-21x^{2}+90x ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&2&-11&-21&90&0\\& & -6& 51& -90& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{30}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-11x^{3}-21x^{2}+90x }{ x+3 } = \color{blue}{2x^{3}-17x^{2}+30x} $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-11&-21&90&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 2 }&-11&-21&90&0\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-11&-21&90&0\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrr}-3&2&\color{orangered}{ -11 }&-21&90&0\\& & \color{orangered}{-6} & & & \\ \hline &2&\color{orangered}{-17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-11&-21&90&0\\& & -6& \color{blue}{51} & & \\ \hline &2&\color{blue}{-17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 51 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}-3&2&-11&\color{orangered}{ -21 }&90&0\\& & -6& \color{orangered}{51} & & \\ \hline &2&-17&\color{orangered}{30}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 30 } = \color{blue}{ -90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-11&-21&90&0\\& & -6& 51& \color{blue}{-90} & \\ \hline &2&-17&\color{blue}{30}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 90 } + \color{orangered}{ \left( -90 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&2&-11&-21&\color{orangered}{ 90 }&0\\& & -6& 51& \color{orangered}{-90} & \\ \hline &2&-17&30&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&2&-11&-21&90&0\\& & -6& 51& -90& \color{blue}{0} \\ \hline &2&-17&30&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&2&-11&-21&90&\color{orangered}{ 0 }\\& & -6& 51& -90& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-17}&\color{blue}{30}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-17x^{2}+30x } $ with a remainder of $ \color{red}{ 0 } $.