Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&16&-20&-4&15\\& & -48& 204& \color{black}{-600} \\ \hline &\color{blue}{16}&\color{blue}{-68}&\color{blue}{200}&\color{orangered}{-585} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x+3 } = \color{blue}{16x^{2}-68x+200} \color{red}{~-~} \frac{ \color{red}{ 585 } }{ x+3 } $$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}{ \left( -48 \right) } = \color{orangered}{ -68 } $
$$ \begin{array}{c|rrrr}-3&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{-48} & & \\ \hline &16&\color{orangered}{-68}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -68 \right) } = \color{blue}{ 204 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&16&-20&-4&15\\& & -48& \color{blue}{204} & \\ \hline &16&\color{blue}{-68}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 204 } = \color{orangered}{ 200 } $
$$ \begin{array}{c|rrrr}-3&16&-20&\color{orangered}{ -4 }&15\\& & -48& \color{orangered}{204} & \\ \hline &16&-68&\color{orangered}{200}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 200 } = \color{blue}{ -600 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&16&-20&-4&15\\& & -48& 204& \color{blue}{-600} \\ \hline &16&-68&\color{blue}{200}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -600 \right) } = \color{orangered}{ -585 } $
$$ \begin{array}{c|rrrr}-3&16&-20&-4&\color{orangered}{ 15 }\\& & -48& 204& \color{orangered}{-600} \\ \hline &\color{blue}{16}&\color{blue}{-68}&\color{blue}{200}&\color{orangered}{-585} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}-68x+200 } $ with a remainder of $ \color{red}{ -585 } $.