Polynomials (1 Viewer)

[khorne]

Is there a better way to do this question? My working is listed below:

P(x) is a monic polynomial of degree 4 with integer coefficients and a constant term 4. One zero is sqrt(2), another zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

Working:

Since (x-sqrt(2)) is a term, and one other term is rational, that means the second term is (x +/- sqrt(2)). The other terms could be +/- 1 and +/- 2.

P(sqrt(2)) = 0, i.e (8+2b) = 0 and 2a + c = 0 ( after subbing and equating)

therefore: p(x) = x^4 + ax^3 -4x^2 -2ax + 4, after factorising: (x^2 -2)(x^2 +ax -2). Taking the quadratic, the roots are: ab = -2, so one set of roots is -1 | 2 or 1 | -2, and additionally, the sign of the sqrt(2) must be positive.

Through trial and error, it can be shown that a = -1 and b = 2 is the correct set, i.e

p(x) = (x- sqrt(2))(x + sqrt(2))(x+1)(x-2), which is the correct answer. My question is...Is there an easier way to do this? Initially I thought just trial and error with some limitations on signs would be faster...

 

[untouchablecuz]

 

[khorne]

Why does -sqrt(2) have to also be a root? I thought conjugates only applied to complex numbers? or does it apply to all irrational roots.

EDIT: Clarification needed

 

[untouchablecuz]

Why does -sqrt(2) have to also be a root? I thought conjugates only applied to complex numbers? or does it apply to all irrational roots.

EDIT: Clarification needed

it's called the irrational conjugate roots theorem (or something along those lines):

if a+b√c is a root of P(x), and:

- √c is irrational
- [a, b] are rational
- the co-efficients of P(x) are real and integral

then a-b√c is also a root of P(x).

google doesn't bring up many results. i can't seem to find/devise a proof for it

someone?

edit: i dont have cambridge 4 unit but it may be in there

 

[khorne]

Yeah, I was a little suspicious of the theorem too since only dr math had a link, but no proof. It seems to work for arbitrary values (n =1,2,3,4)

Edit: Coroneos has it

 

[untouchablecuz]

look @ 2001 MX2 HSC

7. b) runs through the specific case of a cubic (ax³-3x+b=0)

a general proof using the method used here would be way too long

 

[khorne]

yeah, thank you very much, I found a nice additional proof in coro.

For the benefit of the proof:

Assume P(x) has rational co-efficients, and a+sqrt(b) is a zero.

Consider s(x) = {x-(a+sqrt(b)}{x-(a-sqrt(b)} = x^2 -2ax + (a^2 -b)

Divide P(x) by S(x), i.e p(x) = s(x)q(x) +(mx+n), however p(a+sqrt(b)) = 0, s(a+sqrt(b) = 0. From this, it can be shwon that m(a+sqrt(b)+n = 0, msqrt(b) = 0 and ma+n = 0. Thus, m and n = 0

Therefore, s(x) is a factor pf P(x), thus, if (a+sqrt(b)) is a factor, so is (a-sqrt(b))

 

[untouchablecuz]

yeah, thank you very much, I found a nice additional proof in coro.

np. out of curiosity, how does coro prove it?

 

[khorne]

Posted above