polynomials (1 Viewer)

[martin310015]

can someone plz explain to me what does irreducible factors over Q, R and C mean thanx.......... i have no idea what the question asks with this

 

[nike33]

q rational r real c complex

 

[beta-omega]

lemme give ya an example, say the function

f(x)=x^4-9
over the field Q that is all real numbers and no surds
f(x)=(x^2-3)(x^2+3)
since (x^2-3) has a discriminant >0, then it can be simplied with real numbers. However as x= - 3, and 3, then it must only be simplified over the field R, which is all real numbers and irration numbers, ie surds.
f(x)=(x- 3)(x+ 3)(x^2+3)
(x^2+3) has a discriminant of less than zero, so it can only be reduced over the field C, the complex field.
f(x)=(x- 3)(x+ 3)(x-( 3)(i)) (x+( 3)(i))

btw √ means square root

 

[turtle_2468]

basically, reducible means having factors.
The factors in say Q have to have coefficients in Q etc...
ditto R and C.

 

[martin310015]

ok i got a question what happens when u have a polynomial that can not be factorised.........wha tdo i have to do??

 

[CM_Tutor]

All polynomials can be factorised into linear factors over C and all polynomials with real coefficients can be factorised into linear and / or quadratic factors over R.

 

[turtle_2468]

Moved...

 

[KeypadSDM]

Originally posted by martin310015
ok i got a question what happens when u have a polynomial that can not be factorised.........wha tdo i have to do??

Figure out how you factorise it.

 

[Calculon]

Originally posted by martin310015
ok i got a question what happens when u have a polynomial that can not be factorised.........wha tdo i have to do??

Sit there and try stuff til it clicks

 

[+Po1ntDeXt3r+]

wats seems so unsolvable?

cos everything should be solvable into linear factors .. over the Complex roots

 

[turtle_2468]

lol.. should.
doesn't mean there's a nice way to do it

 

[+Po1ntDeXt3r+]

pfft.... how bad could it be? (im feeling cocky)
Post the worse things up.. BRING IT ON

also
turtle? arent u in med II yet?
not tat repeatin is a bad thing

 

[turtle_2468]

changed.

 

[Mill]

Martin, if you haven't done the question yet:

(i) Q : rational
(ii) R : real
(iii) C : complex

Have a look at your notes from week 1; in the first few pages you put some numbers like 2 and pi and 1 + i into a diagram that sort of tried to categorise some numbers as rational, real, complex, etc.

So, for part (i) you need to factorise as far as possible so that you have no surds and no complex numbers. In part (ii), you want no complex numbers, but you can have surds. And in part (iii), you want both surds and complex numbers.

 

[nike33]

"wats seems so unsolvable?

cos everything should be solvable into linear factors .. over the Complex roots"
"

hehe try solving this one x⁵ + Ax⁴+Bx³ + Cx²+Dx + C

 

[Grey Council]

why's this in the extracurricular?

isn't this in 4u polynomials?
hrm, maybe i don't understand this
hrm

 

[Mill]

It is normal 4u. It's from a 4u worksheet. So I'm not sure why it's been moved

 

[CM_Tutor]

I think there was a long buchanan post that was extra-curricular, that was just after mine and just before turtle_2468's. That was (I think) what prompted the move.

 

[turtle_2468]

Yeah... that's about right. only 1% of this post is relevant anyway, the other stuff only serves to confuse ppl...

 

[+Po1ntDeXt3r+]

Originally posted by nike33
hehe try solving this one x⁵ + Ax⁴+Bx³ + Cx²+Dx + C

ok .. (x-i)(x-i)(x-i)(x-i)(x-i) where ,,,, E C
.. im pretti sure thats a soln..