Divide using synthetic division $$ ( 5x^{3}+11x^{2}-14x-6 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&5&11&-14&-6\\& & -15& 12& \color{black}{6} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+11x^{2}-14x-6 }{ x+3 } = \color{blue}{5x^{2}-4x-2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&11&-14&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 5 }&11&-14&-6\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&11&-14&-6\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&5&\color{orangered}{ 11 }&-14&-6\\& & \color{orangered}{-15} & & \\ \hline &5&\color{orangered}{-4}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&11&-14&-6\\& & -15& \color{blue}{12} & \\ \hline &5&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&5&11&\color{orangered}{ -14 }&-6\\& & -15& \color{orangered}{12} & \\ \hline &5&-4&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&11&-14&-6\\& & -15& 12& \color{blue}{6} \\ \hline &5&-4&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-3&5&11&-14&\color{orangered}{ -6 }\\& & -15& 12& \color{orangered}{6} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-4x-2 } $ with a remainder of $ \color{red}{ 0 } $.