Divide using synthetic division $$ ( 16x^{3}-20x^{2}-4x+15 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&16&-20&-4&15\\& & 64& 176& \color{black}{688} \\ \hline &\color{blue}{16}&\color{blue}{44}&\color{blue}{172}&\color{orangered}{703} \end{array} $$The solution is:
$$ \frac{ 16x^{3}-20x^{2}-4x+15 }{ x-4 } = \color{blue}{16x^{2}+44x+172} ~+~ \frac{ \color{red}{ 703 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&16&-20&-4&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 16 }&-20&-4&15\\& & & & \\ \hline &\color{orangered}{16}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 16 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&16&-20&-4&15\\& & \color{blue}{64} & & \\ \hline &\color{blue}{16}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 64 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrr}4&16&\color{orangered}{ -20 }&-4&15\\& & \color{orangered}{64} & & \\ \hline &16&\color{orangered}{44}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 44 } = \color{blue}{ 176 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&16&-20&-4&15\\& & 64& \color{blue}{176} & \\ \hline &16&\color{blue}{44}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 176 } = \color{orangered}{ 172 } $
$$ \begin{array}{c|rrrr}4&16&-20&\color{orangered}{ -4 }&15\\& & 64& \color{orangered}{176} & \\ \hline &16&44&\color{orangered}{172}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 172 } = \color{blue}{ 688 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&16&-20&-4&15\\& & 64& 176& \color{blue}{688} \\ \hline &16&44&\color{blue}{172}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 688 } = \color{orangered}{ 703 } $
$$ \begin{array}{c|rrrr}4&16&-20&-4&\color{orangered}{ 15 }\\& & 64& 176& \color{orangered}{688} \\ \hline &\color{blue}{16}&\color{blue}{44}&\color{blue}{172}&\color{orangered}{703} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{2}+44x+172 } $ with a remainder of $ \color{red}{ 703 } $.