Divide using synthetic division $$ ( 15x^{3}+34x^{2}-25 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&15&34&0&-25\\& & 15& 49& \color{black}{49} \\ \hline &\color{blue}{15}&\color{blue}{49}&\color{blue}{49}&\color{orangered}{24} \end{array} $$The solution is:
$$ \frac{ 15x^{3}+34x^{2}-25 }{ x-1 } = \color{blue}{15x^{2}+49x+49} ~+~ \frac{ \color{red}{ 24 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&34&0&-25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 15 }&34&0&-25\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 15 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&34&0&-25\\& & \color{blue}{15} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ 15 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}1&15&\color{orangered}{ 34 }&0&-25\\& & \color{orangered}{15} & & \\ \hline &15&\color{orangered}{49}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 49 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&34&0&-25\\& & 15& \color{blue}{49} & \\ \hline &15&\color{blue}{49}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 49 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}1&15&34&\color{orangered}{ 0 }&-25\\& & 15& \color{orangered}{49} & \\ \hline &15&49&\color{orangered}{49}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 49 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&15&34&0&-25\\& & 15& 49& \color{blue}{49} \\ \hline &15&49&\color{blue}{49}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 49 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}1&15&34&0&\color{orangered}{ -25 }\\& & 15& 49& \color{orangered}{49} \\ \hline &\color{blue}{15}&\color{blue}{49}&\color{blue}{49}&\color{orangered}{24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}+49x+49 } $ with a remainder of $ \color{red}{ 24 } $.