Divide using synthetic division $$ ( 4x^{3}-14x^{2}+4x+15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-14&4&15\\& & 12& -6& \color{black}{-6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-14x^{2}+4x+15 }{ x-3 } = \color{blue}{4x^{2}-2x-2} ~+~ \frac{ \color{red}{ 9 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-14&4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-14&4&15\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-14&4&15\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -14 }&4&15\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-14&4&15\\& & 12& \color{blue}{-6} & \\ \hline &4&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}3&4&-14&\color{orangered}{ 4 }&15\\& & 12& \color{orangered}{-6} & \\ \hline &4&-2&\color{orangered}{-2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-14&4&15\\& & 12& -6& \color{blue}{-6} \\ \hline &4&-2&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}3&4&-14&4&\color{orangered}{ 15 }\\& & 12& -6& \color{orangered}{-6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-2x-2 } $ with a remainder of $ \color{red}{ 9 } $.