Divide using synthetic division $$ ( 15x^{3}-89x^{2}+21x+245 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&15&-89&21&245\\& & 75& -70& \color{black}{-245} \\ \hline &\color{blue}{15}&\color{blue}{-14}&\color{blue}{-49}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 15x^{3}-89x^{2}+21x+245 }{ x-5 } = \color{blue}{15x^{2}-14x-49} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&15&-89&21&245\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 15 }&-89&21&245\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 15 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&15&-89&21&245\\& & \color{blue}{75} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -89 } + \color{orangered}{ 75 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}5&15&\color{orangered}{ -89 }&21&245\\& & \color{orangered}{75} & & \\ \hline &15&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -70 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&15&-89&21&245\\& & 75& \color{blue}{-70} & \\ \hline &15&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -70 \right) } = \color{orangered}{ -49 } $
$$ \begin{array}{c|rrrr}5&15&-89&\color{orangered}{ 21 }&245\\& & 75& \color{orangered}{-70} & \\ \hline &15&-14&\color{orangered}{-49}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -49 \right) } = \color{blue}{ -245 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&15&-89&21&245\\& & 75& -70& \color{blue}{-245} \\ \hline &15&-14&\color{blue}{-49}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 245 } + \color{orangered}{ \left( -245 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&15&-89&21&\color{orangered}{ 245 }\\& & 75& -70& \color{orangered}{-245} \\ \hline &\color{blue}{15}&\color{blue}{-14}&\color{blue}{-49}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}-14x-49 } $ with a remainder of $ \color{red}{ 0 } $.