Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&16&-20&-4&15\\& & 80& 300& \color{black}{1480} \\ \hline &\color{blue}{16}&\color{blue}{60}&\color{blue}{296}&\color{orangered}{1495} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-5 } = \color{blue}{16x^{2}+60x+296} ~+~ \frac{ \color{red}{ 1495 } }{ x-5 } $$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}{ 80 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}5&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{80} & & \\ \hline &16&\color{orangered}{60}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 60 } = \color{blue}{ 300 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&16&-20&-4&15\\& & 80& \color{blue}{300} & \\ \hline &16&\color{blue}{60}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 300 } = \color{orangered}{ 296 } $
$$ \begin{array}{c|rrrr}5&16&-20&\color{orangered}{ -4 }&15\\& & 80& \color{orangered}{300} & \\ \hline &16&60&\color{orangered}{296}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 296 } = \color{blue}{ 1480 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&16&-20&-4&15\\& & 80& 300& \color{blue}{1480} \\ \hline &16&60&\color{blue}{296}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 1480 } = \color{orangered}{ 1495 } $
$$ \begin{array}{c|rrrr}5&16&-20&-4&\color{orangered}{ 15 }\\& & 80& 300& \color{orangered}{1480} \\ \hline &\color{blue}{16}&\color{blue}{60}&\color{blue}{296}&\color{orangered}{1495} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+60x+296 } $ with a remainder of $ \color{red}{ 1495 } $.