GCD of f(x) and f'(x) (simple question)

• Nov 25th 2009, 12:38 PM
elninio
GCD of f(x) and f'(x) (simple question)
This seemed simple, but I keep getting ridiculous fractions.

For example, find the GCD of f(x) and f'(x) over Q of:
f(x)=x^3 -3x -2

The quotient becomes 1/3x with a remainder of -2x-2, which can be simplified to x+1.
From my understanding, x+1 is the GCD. This doesnt seem correct, taking the quotient into consideration. Can anyone check this?
As a sample, my book has GCD(f(x),f'(x))= x-1 when f(x)=x^4-x^3-x+1. I could not work this question out!

Also, as a very similar question, if I get a remainder of -2x^2+x, can I simplify this to -2x+1, or does such a simplification not work for GCD's?
• Nov 25th 2009, 01:13 PM
tonio
Quote:

Originally Posted by elninio
This seemed simple, but I keep getting ridiculous fractions.

For example, find the GCD of f(x) and f'(x) over Q of:
f(x)=x^3 -3x -2

The quotient becomes 1/3x with a remainder of -2x-2, which can be simplified to x+1.
From my understanding, x+1 is the GCD. This doesnt seem correct,

But it is: just check that both $\displaystyle f(x)\,,\,f'(x)$ have -1 as one of their roots.

taking the quotient into consideration. Can anyone check this?

Also, as a very similar question, if I get a remainder of -2x^2+x, can I simplify this to -2x+1, or does such a simplification not work for GCD's?

What do you mean "simplify it"? You're dividing it by a non-trivial divisor! That's not simplifying but changing.

Perhpas what you mean is that, for instance, $\displaystyle 2x+2\,,\,x+1$ are associate or equivalent elements since one equals the other times a unit.
In your case, if $\displaystyle -2x^2+x=-x(2x-1)=\frac{1}{2}x\!\!\left(x-\frac{1}{2}\right)$ is the gcd, then you can "factor out" that unit and say the gcd is $\displaystyle x\left(x-\frac{1}{2}\right)$

Tonio

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• Nov 25th 2009, 01:15 PM
elninio
Thanks Tonio. That cleared everything up. You're the man!
• Nov 25th 2009, 01:23 PM
elninio
I'm not sure if I need to make a new thread, but could you check one final question:

GCD$\displaystyle (x^3-2x^2+3x+1,x^3+2x+1)$ over $\displaystyle Z_5$.

I worked it out and got a quotient of 1 with remainder $\displaystyle 2x^2+x (=x(2x+1))$
• Nov 25th 2009, 01:30 PM
tonio
Quote:

Originally Posted by elninio
I'm not sure if I need to make a new thread, but could you check one final question:

GCD$\displaystyle (x^3-2x^2+3x+1,x^3+2x+1)$ over $\displaystyle Z_5$.

I worked it out and got a quotient of 1 with remainder $\displaystyle 2x^2+x (=x(2x+1))$

Well, the quotient is correct but the remainder I got is $\displaystyle 3x^2+x$...and now what? You haven't yet ginished calculating the GCD of those two pol's!
I advice you to re-read Euclides Method to determine the gcd of two elements in an Euclidean Domain.

Tonio
• Nov 25th 2009, 01:51 PM
elninio
Ahhh, I see. Then I would divide the lower degree polynomial by the remainder until I find a remainder of zero then, correct?
• Nov 25th 2009, 02:11 PM
tonio
Quote:

Originally Posted by elninio
Ahhh, I see. Then I would divide the lower degree polynomial by the remainder until I find a remainder of zero then, correct?

Yep, so...go on!

Tonio
• Nov 25th 2009, 02:32 PM
elninio
(Disregard this...>SOLVED)