Again, from a guy at work...this question has me stumped:
(4x^3+x+10) / (2x-1)
The answer in the book is 2x^2+x+1+11/2x-1
I for the life of me can not figure out how they are getting the +1+11/2x-1
This is done by long division.
$\displaystyle \frac{4x^3 + x + 10}{2x - 1} = \frac{4x^3 + 0\cdot x^2 + x + 10}{2x - 1}$
I'm not going to write this out showing all the work, I'll just do it this way:
$\displaystyle \frac{4x^3 + 0\cdot x^2 + x + 10}{2x - 1}$
$\displaystyle = \frac{(4x^3 - 2x^2) + 2x^2 + x + 10}{2x - 1} = \frac{4x^3 - 2x^2}{2x - 1} + \frac{2x^2 + x + 10}{2x - 1}$
$\displaystyle = 2x^2 + \frac{2x^2 + x + 10}{2x - 1}$
$\displaystyle = 2x^2 + \frac{(2x^2 - x) + x + x + 10}{2x - 1} = 2x^2 + \frac{2x^2 - x}{2x - 1} + \frac{2x + 10}{2x - 1}$
$\displaystyle = 2x^2 + x + \frac{2x + 10}{2x - 1}$
$\displaystyle = 2x^2 + x + \frac{(2x - 1) + 1 + 10}{2x - 1} = 2x^2 + x + \frac{2x - 1}{2x - 1} + \frac{11}{2x - 1}$
$\displaystyle = 2x^2 + x + 1 + \frac{11}{2x - 1}$
-Dan