Divide using synthetic division $$ ( 3x^{3}-14x^{2}-12x+7 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&3&-14&-12&7\\& & 15& 5& \color{black}{-35} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-7}&\color{orangered}{-28} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-14x^{2}-12x+7 }{ x-5 } = \color{blue}{3x^{2}+x-7} \color{red}{~-~} \frac{ \color{red}{ 28 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-14&-12&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 3 }&-14&-12&7\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-14&-12&7\\& & \color{blue}{15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 15 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&3&\color{orangered}{ -14 }&-12&7\\& & \color{orangered}{15} & & \\ \hline &3&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-14&-12&7\\& & 15& \color{blue}{5} & \\ \hline &3&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 5 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}5&3&-14&\color{orangered}{ -12 }&7\\& & 15& \color{orangered}{5} & \\ \hline &3&1&\color{orangered}{-7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-14&-12&7\\& & 15& 5& \color{blue}{-35} \\ \hline &3&1&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrr}5&3&-14&-12&\color{orangered}{ 7 }\\& & 15& 5& \color{orangered}{-35} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{-7}&\color{orangered}{-28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+x-7 } $ with a remainder of $ \color{red}{ -28 } $.