1. ## dividing polynomials

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

2. They used the remainder. The remainder is $\frac{11}{2x-1}$. Also $2x-1$ goes into $2x+10$ once. And thats how we get the remainder term.

3. Originally Posted by Stuck686
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.
$\frac{4x^3 + x + 10}{2x - 1} = \frac{4x^3 + 0\cdot x^2 + x + 10}{2x - 1}$

I'm not going to write this out showing all the work, I'll just do it this way:
$\frac{4x^3 + 0\cdot x^2 + x + 10}{2x - 1}$

$= \frac{(4x^3 - 2x^2) + 2x^2 + x + 10}{2x - 1} = \frac{4x^3 - 2x^2}{2x - 1} + \frac{2x^2 + x + 10}{2x - 1}$

$= 2x^2 + \frac{2x^2 + x + 10}{2x - 1}$

$= 2x^2 + \frac{(2x^2 - x) + x + x + 10}{2x - 1} = 2x^2 + \frac{2x^2 - x}{2x - 1} + \frac{2x + 10}{2x - 1}$

$= 2x^2 + x + \frac{2x + 10}{2x - 1}$

$= 2x^2 + x + \frac{(2x - 1) + 1 + 10}{2x - 1} = 2x^2 + x + \frac{2x - 1}{2x - 1} + \frac{11}{2x - 1}$

$= 2x^2 + x + 1 + \frac{11}{2x - 1}$

-Dan

4. Originally Posted by Stuck686
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
Hello,

I assume that you know how to do synthetic division(?):
Code:
                                                   11
( 4x³ + 0*x² + x + 10) ÷ (2x - 1) = 2x² + x + 1 + ------
( 4x³- 2x²)                                       2x-1
-----------
2x² +  x
(2x² - x )
-----------
2x + 10
(2x -  1)
---------
11
You have to list all powers of x in the dividend.