Divide using synthetic division $$ ( 5x^{3}-6x^{2}+15 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&5&-6&0&15\\& & -10& 32& \color{black}{-64} \\ \hline &\color{blue}{5}&\color{blue}{-16}&\color{blue}{32}&\color{orangered}{-49} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-6x^{2}+15 }{ x+2 } = \color{blue}{5x^{2}-16x+32} \color{red}{~-~} \frac{ \color{red}{ 49 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-6&0&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 5 }&-6&0&15\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-6&0&15\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-2&5&\color{orangered}{ -6 }&0&15\\& & \color{orangered}{-10} & & \\ \hline &5&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-6&0&15\\& & -10& \color{blue}{32} & \\ \hline &5&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 32 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}-2&5&-6&\color{orangered}{ 0 }&15\\& & -10& \color{orangered}{32} & \\ \hline &5&-16&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 32 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-6&0&15\\& & -10& 32& \color{blue}{-64} \\ \hline &5&-16&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -49 } $
$$ \begin{array}{c|rrrr}-2&5&-6&0&\color{orangered}{ 15 }\\& & -10& 32& \color{orangered}{-64} \\ \hline &\color{blue}{5}&\color{blue}{-16}&\color{blue}{32}&\color{orangered}{-49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-16x+32 } $ with a remainder of $ \color{red}{ -49 } $.