Divide using synthetic division $$ ( 3x^{3}-17x^{2}-2x+12 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&3&-17&-2&12\\& & 15& -10& \color{black}{-60} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-12}&\color{orangered}{-48} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-17x^{2}-2x+12 }{ x-5 } = \color{blue}{3x^{2}-2x-12} \color{red}{~-~} \frac{ \color{red}{ 48 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-17&-2&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 3 }&-17&-2&12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-17&-2&12\\& & \color{blue}{15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 15 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}5&3&\color{orangered}{ -17 }&-2&12\\& & \color{orangered}{15} & & \\ \hline &3&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-17&-2&12\\& & 15& \color{blue}{-10} & \\ \hline &3&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}5&3&-17&\color{orangered}{ -2 }&12\\& & 15& \color{orangered}{-10} & \\ \hline &3&-2&\color{orangered}{-12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-17&-2&12\\& & 15& -10& \color{blue}{-60} \\ \hline &3&-2&\color{blue}{-12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -48 } $
$$ \begin{array}{c|rrrr}5&3&-17&-2&\color{orangered}{ 12 }\\& & 15& -10& \color{orangered}{-60} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-12}&\color{orangered}{-48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-2x-12 } $ with a remainder of $ \color{red}{ -48 } $.