Divide using synthetic division $$ ( 5x^{4}+11x^{3}-10x^{2}+2x-6 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&5&11&-10&2&-6\\& & -15& 12& -6& \color{black}{12} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+11x^{3}-10x^{2}+2x-6 }{ x+3 } = \color{blue}{5x^{3}-4x^{2}+2x-4} ~+~ \frac{ \color{red}{ 6 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&-10&2&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 5 }&11&-10&2&-6\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&-10&2&-6\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&5&\color{orangered}{ 11 }&-10&2&-6\\& & \color{orangered}{-15} & & & \\ \hline &5&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&-10&2&-6\\& & -15& \color{blue}{12} & & \\ \hline &5&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&5&11&\color{orangered}{ -10 }&2&-6\\& & -15& \color{orangered}{12} & & \\ \hline &5&-4&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&-10&2&-6\\& & -15& 12& \color{blue}{-6} & \\ \hline &5&-4&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&5&11&-10&\color{orangered}{ 2 }&-6\\& & -15& 12& \color{orangered}{-6} & \\ \hline &5&-4&2&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&11&-10&2&-6\\& & -15& 12& -6& \color{blue}{12} \\ \hline &5&-4&2&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 12 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-3&5&11&-10&2&\color{orangered}{ -6 }\\& & -15& 12& -6& \color{orangered}{12} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-4x^{2}+2x-4 } $ with a remainder of $ \color{red}{ 6 } $.