The Student Room Group

coprime polynomials

I keep getting the wrong answer...

Show that the polynomials

f(x) = x^4 + 3x^3 - 2x^2 - x - 1
g(x) = 2x^2 + 2x + 1

are coprime.

I've already got that

x^4 + 3x^3 - 2x^2 - x - 1 = (0.5x^2 + x - 2.25)*(2x^2 + 2x + 1) + (2.5x + 1.25)

I've been trying to use Euclid's Algorithm, but I'm not really sure where I'm going wrong (although I clearly am since I can't the highest common factor = 1).

Thanks.
Reply 1
Two polynomials are coprime iff they share no roots.

Suppose that x is such that

x^4 + 3x^3 - 2x^2 - x - 1 = 2x^2 + 2x + 1 = 0

Then, using your division, 2.5x + 1.25 = 0, so x = -1/2. But x = -1/2 is not a root of either polynomial. Contradiction.
flyinghorse
I keep getting the wrong answer...

Show that the polynomials

f(x) = x^4 + 3x^3 - 2x^2 - x - 1
g(x) = 2x^2 + 2x + 1

are coprime.

Thanks.

conside g(x)=0
x=-2+sqrt(4-8)/4
so over the reals g(x) is irreducible.
since g(x) is not a factor of f(x), as shown in your post, the result follows.
Reply 3
evariste
conside g(x)=0
x=-2+sqrt(4-8)/4
so over the reals g(x) is irreducible.
since g(x) is not a factor of f(x), as shown in your post, the result follows.

Isn't the hypothetical common factor allowed to be complex?
Jonny W
Isn't the hypothetical common factor allowed to be complex?


maybe,but normally these questions are set over the reals. if this over C then they should have said!!
Reply 5
thanks for your help

I don't suppose you could help me with finding a,b (polynomials) s.t.

a(x)f(x) + b(x)g(x) = 1?
Reply 6
flyinghorse


I've already got that

x^4 + 3x^3 - 2x^2 - x - 1 = (0.5x^2 + x - 2.25)*(2x^2 + 2x + 1) + (2.5x + 1.25)

I've been trying to use Euclid's Algorithm, but I'm not really sure where I'm going wrong (although I clearly am since I can't the highest common factor = 1).

Thanks.


You've only applied what's called the division algorithm once. The Euclidean algorithm requires you to keep going till you get 0, and then the previous term is the hcf.

So divide 2.5x+1.25 into 2x^2 + 2x + 1. See what remainder you get.

Then if it's non-zero divide the remainder into 2.5x+1.25 etc.
Reply 7
RichE
You've only applied what's called the division algorithm once. The Euclidean algorithm requires you to keep going till you get 0, and then the previous term is the hcf.

So divide 2.5x+1.25 into 2x^2 + 2x + 1. See what remainder you get.

Then if it's non-zero divide the remainder into 2.5x+1.25 etc.


I've applied Euclid's Algorithm repeatedly, but at one point I get a remainder of (x+1), then divide by that and get a negative remainder, so I'm not sure where I've gone wrong.
flyinghorse
thanks for your help

I don't suppose you could help me with finding a,b (polynomials) s.t.

a(x)f(x) + b(x)g(x) = 1?

hard to type in!!
x^4+3x^3-2x^2-x+1=(x^2/2+x-9/4)(2x^2+2x+1)+5x/2+13/4
2x^2+2x+1=(5x/2+13/4)(4x/5-6/25)-178/100
going back gives
((x^2/2+x-9/4)(2x^2+2x+1)-(x^4+3x^3-2x^2-x+1))(4x/5-6/25)+(2x^2+2x+1)=178/100
(2x^2+2x+1)((x^2/2+x-9/4)(4x/5-6/25)+1)-(x^4+3x^3-2x^2-x+1)(4x/5-6/25)=178/100
grrrr. wrote down the wrong eqn.
lets try again ive pasted your post to ensure i get correct f,g
x^4 + 3x^3 - 2x^2 - x - 1 = (0.5x^2 + x - 2.25)*(2x^2 + 2x + 1) + (2.5x + 1.25)
=(x^2/2+x-9/4)(2x^2+2x+1)+(5x/2+5/4)
2x^2+2x+1=(5x/2+5/4)(4x/5+2/5)+1/2
1/2=2x^2+2x+1-(5x/2+5/4)(4x/5+2/5)
=2x^2+2x+1-(-(x^2/2+x-9/4)(2x^2+2x+1)+(x^4 + 3x^3 - 2x^2 - x - 1))(4x/5+2/5)
so
1/2=(2x^2+2x+1)(1+(x^2/2+x-9/4)(4x/5+2/5))-(x^4 + 3x^3 - 2x^2 - x - 1)(4x/5+2/5)
result follows