Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&16&-20&-4&15\\& & 48& 84& \color{black}{240} \\ \hline &\color{blue}{16}&\color{blue}{28}&\color{blue}{80}&\color{orangered}{255} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-3 } = \color{blue}{16x^{2}+28x+80} ~+~ \frac{ \color{red}{ 255 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&16&-20&-4&15\\& & \color{blue}{48} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 48 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}3&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{48} & & \\ \hline &16&\color{orangered}{28}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 28 } = \color{blue}{ 84 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&16&-20&-4&15\\& & 48& \color{blue}{84} & \\ \hline &16&\color{blue}{28}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 84 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrr}3&16&-20&\color{orangered}{ -4 }&15\\& & 48& \color{orangered}{84} & \\ \hline &16&28&\color{orangered}{80}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 80 } = \color{blue}{ 240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&16&-20&-4&15\\& & 48& 84& \color{blue}{240} \\ \hline &16&28&\color{blue}{80}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 240 } = \color{orangered}{ 255 } $
$$ \begin{array}{c|rrrr}3&16&-20&-4&\color{orangered}{ 15 }\\& & 48& 84& \color{orangered}{240} \\ \hline &\color{blue}{16}&\color{blue}{28}&\color{blue}{80}&\color{orangered}{255} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+28x+80 } $ with a remainder of $ \color{red}{ 255 } $.