Divide using synthetic division $$ ( 3x^{3}-14x^{2}+16x+2 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&-14&16&2\\& & 9& -15& \color{black}{3} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{1}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-14x^{2}+16x+2 }{ x-3 } = \color{blue}{3x^{2}-5x+1} ~+~ \frac{ \color{red}{ 5 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-14&16&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&-14&16&2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-14&16&2\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 9 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ -14 }&16&2\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-14&16&2\\& & 9& \color{blue}{-15} & \\ \hline &3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&3&-14&\color{orangered}{ 16 }&2\\& & 9& \color{orangered}{-15} & \\ \hline &3&-5&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-14&16&2\\& & 9& -15& \color{blue}{3} \\ \hline &3&-5&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&3&-14&16&\color{orangered}{ 2 }\\& & 9& -15& \color{orangered}{3} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{1}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-5x+1 } $ with a remainder of $ \color{red}{ 5 } $.