# Algebraic Fractions question

• Oct 3rd 2006, 11:12 AM
neo658
Algebraic Fractions question
I'v been stuck on this question and I dont know how to solve it. I have to use the remainder theorem to find the answer for (4x³ - 5x² + 3x - 14) / (x² + 2x - 1).

Since I can't expand (x² + 2x - 1) so I cannot find any values for "x". Can somebody please help I have to hand this in after tomorrow.

Thanks
• Oct 3rd 2006, 11:21 AM
ThePerfectHacker
Quote:

Originally Posted by neo658
I'v been stuck on this question and I dont know how to solve it. I have to use the remainder theorem to find the answer for (4x&#179; - 5x&#178; + 3x - 14) / (x&#178; + 2x - 1).

Since I can't expand (x&#178; + 2x - 1) so I cannot find any values for "x". Can somebody please help I have to hand this in after tomorrow.

Thanks

Code:

``` 4x^3-5x^2+3x-14 -:- x^2+2x-1=4x-13 -4x^3-8x^2+4x ------------------ -13x^2+ 7x  -14 +13x^2+26x-13 -------------------- 33x-27 (this is the remainder)```
This is mine 27:):)th Post!!!
• Oct 3rd 2006, 12:18 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Code:

``` 4x^3-5x^2+3x-14 -:- x^2+2x-1=4x-13 -4x^3-8x^2+4x ------------------ -13x^2+ 7x  -14 +13x^2+26x-13 -------------------- 33x-27 (this is the remainder)```
This is mine 27:):)th Post!!!

To expand on TPH's answer a little bit it is based on the remainder theorem:
Given polynomials a and b, if the remainder of a/b is r, then a = b*q + r, where q and r are polynomials and the degree of r is 0 <= r < b.

Here a = 4x^3-5x^2+3x-14, b = x^2+2x-1, q = 4x + 13, and r = 33x - 27.

-Dan
• Oct 3rd 2006, 01:45 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
To expand on TPH's answer a little bit it is based on the remainder theorem:
Given polynomials a and b, if the remainder of a/b is r, then a = b*q + r, where q and r are polynomials and the degree of r is 0 <= r < b.

Here a = 4x^3-5x^2+3x-14, b = x^2+2x-1, q = 4x + 13, and r = 33x - 27.

Let me say it even better.
Given polynomials f(x) and g(x) then there exists unique polynomials q(x) and r(x) with deg (r)<deg(g) such that,
f(x)=q(x)f(x)+r(x)

[Existence/Uniqueness happens to be of paramount importance in field theory]
• Oct 4th 2006, 12:29 PM
neo658
uh guys can ya please make this like simpler I dint understandu guys at all :confused:
• Oct 4th 2006, 01:20 PM
ThePerfectHacker
Quote:

Originally Posted by neo658
uh guys can ya please make this like simpler I dint understandu guys at all :confused:

topsquark is to blame.
---
Look at my first post.
Look at how I did long division, and understand it.
The number on the bottom which is left you cannot divide futher.
That number is the "remainder".
The number after the equal sign is the "quotient".
Now you have the quotient and remainder after division.
• Oct 4th 2006, 04:41 PM
topsquark
:o I'm sowwy! :o

-Dan
• Oct 8th 2006, 11:18 AM
neo658
thanks loads guys now so much clearer =)