Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&16&-20&-4&15\\& & -80& 500& \color{black}{-2480} \\ \hline &\color{blue}{16}&\color{blue}{-100}&\color{blue}{496}&\color{orangered}{-2465} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x+5 } = \color{blue}{16x^{2}-100x+496} \color{red}{~-~} \frac{ \color{red}{ 2465 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 16 } = \color{blue}{ -80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&-20&-4&15\\& & \color{blue}{-80} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -80 \right) } = \color{orangered}{ -100 } $
$$ \begin{array}{c|rrrr}-5&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{-80} & & \\ \hline &16&\color{orangered}{-100}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -100 \right) } = \color{blue}{ 500 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&-20&-4&15\\& & -80& \color{blue}{500} & \\ \hline &16&\color{blue}{-100}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 500 } = \color{orangered}{ 496 } $
$$ \begin{array}{c|rrrr}-5&16&-20&\color{orangered}{ -4 }&15\\& & -80& \color{orangered}{500} & \\ \hline &16&-100&\color{orangered}{496}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 496 } = \color{blue}{ -2480 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&16&-20&-4&15\\& & -80& 500& \color{blue}{-2480} \\ \hline &16&-100&\color{blue}{496}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -2480 \right) } = \color{orangered}{ -2465 } $
$$ \begin{array}{c|rrrr}-5&16&-20&-4&\color{orangered}{ 15 }\\& & -80& 500& \color{orangered}{-2480} \\ \hline &\color{blue}{16}&\color{blue}{-100}&\color{blue}{496}&\color{orangered}{-2465} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}-100x+496 } $ with a remainder of $ \color{red}{ -2465 } $.