Divide using synthetic division $$ ( 3x^{3}-10x^{2}-15x+1 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&3&-10&-15&1\\& & 15& 25& \color{black}{50} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{10}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-10x^{2}-15x+1 }{ x-5 } = \color{blue}{3x^{2}+5x+10} ~+~ \frac{ \color{red}{ 51 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-10&-15&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 3 }&-10&-15&1\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-10&-15&1\\& & \color{blue}{15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 15 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}5&3&\color{orangered}{ -10 }&-15&1\\& & \color{orangered}{15} & & \\ \hline &3&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-10&-15&1\\& & 15& \color{blue}{25} & \\ \hline &3&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 25 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}5&3&-10&\color{orangered}{ -15 }&1\\& & 15& \color{orangered}{25} & \\ \hline &3&5&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 10 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-10&-15&1\\& & 15& 25& \color{blue}{50} \\ \hline &3&5&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 50 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrr}5&3&-10&-15&\color{orangered}{ 1 }\\& & 15& 25& \color{orangered}{50} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{10}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+5x+10 } $ with a remainder of $ \color{red}{ 51 } $.