Divide using synthetic division $$ ( 2x^{3}-14x+10 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&0&-14&10\\& & -6& 18& \color{black}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-14x+10 }{ x+3 } = \color{blue}{2x^{2}-6x+4} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&0&-14&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&0&-14&10\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&0&-14&10\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 0 }&-14&10\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&0&-14&10\\& & -6& \color{blue}{18} & \\ \hline &2&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 18 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-3&2&0&\color{orangered}{ -14 }&10\\& & -6& \color{orangered}{18} & \\ \hline &2&-6&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&0&-14&10\\& & -6& 18& \color{blue}{-12} \\ \hline &2&-6&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-3&2&0&-14&\color{orangered}{ 10 }\\& & -6& 18& \color{orangered}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-6x+4 } $ with a remainder of $ \color{red}{ -2 } $.