Harder Polynomial (1 Viewer)

[OLDMAN]

P(x) is a polynomial with integral coefficients. The leading coefficient, the constant term, and P(1) are all odd. Show that P(x) has no rational roots.

 

[Affinity]

nice question

 

[McLake]

Discrim = sqrt (b^2 - 4ac)
b is odd, so b^2 is odd
a and c are odd, so 4ac is even.
odd - even can't be zero so 2 or no roots.

hmm ...

 

[spice girl]

anyone want the solution to this one yet?

i thought of one using mod-arithmetic...i dun think they teach that for 4u...

 

[OLDMAN]

____________________________________________
Spice Girl Quote:
i thought of one using mod-arithmetic...i dun think they teach that for 4u...
___________________________________________

I could hear Spice Girl chuckling already.

I didn't really want to post a non-4U question, so the following should make it a fair q7 type problem.

a) If a_1+a_2+...+a_n is an odd number, prove that
a_1*b_1+a_2*b_2+...+a_n*b_n is an odd number if each b_i is an odd number.

b)P(x) is a polynomial with integral coefficients. The leading coefficient, the constant term, and P(1) are all odd. Show that P(x) has no rational roots. You can use part a).

 

[Affinity]

spice girl: Does modulo arithmetic make it easier? tried that, didn't come out nicely.

 

[underthesun]

excuse my ignorance, but what's integral co-efficients?
last year I wagged most complex number and polynomial classes, and this year I joined the class late..

 

[Affinity]

underthesun: all coefficients are whole numbers.

like x^3 + 1023x^2 -4x + (-2)

 

[wogboy]

Nice proof Affinity.

 

[spice girl]

hell yeh it comes out nicely...

suppose x=p/q is a solution
then by rational root thm, since both leading coefficient and constant term is odd, p and q must both be odd (p | a_0, q | a_n)

So P(p/q) = 0
=> a_n*p^n/q^n + a_n-1*p^n-1/q^n-1 + ... + a_1*p/q + a_0 = 0
=> a_n*p^n + a_n-1*p^n-1*q + ... + a_1*p*q^n-1 + a_0*q^n = 0

(notation ~= is "congruent to")
So P(p/q) ~= a_n*(1) + a_n-1*(1) + ... + a_1*(1) + a_0*(1) (mod 2)
(since p, q are both odd)
~= (1)*(a_n + a_n-1 + ... + a_1 + a_0)
~= (1)*P(1)
~= 1
so P(p/q) is an odd integer which is certainly not = 0
thus contradiction.

 

[OLDMAN]

Never thought it will be done this quickly -two solutions at that on an unfair question. If this was a real fair q7, it would have an additional lead up: i) Let x=p/q, a root, where p and q are integers having no common divisor...explain why p divides a_n, and q divides a_0.

Now Affinity :take a leisurely stroll through 2001 Q7 (b).

 

[Affinity]

Thanx spice girl. that was short~
damn, overlooked the property that p| a_n and q| a_0
and took a long way

And thanks to OLDMAN too

 

[fatmuscle]

nice work!!!

very nice Q

 

[ND]

spice girl: did you learn that stuff before uni? If so was it at all helpful for the hsc?

 

[spice girl]

no usually we're not allowed to introduce too much weird new stuff into our HSC 4u, but i learnt this in yr7 maths enrichment...surprising NSW doesn't teach it at all...

 

[Affinity]

surprising NSW doesn't teach it at all...

Do they teach it in some other state?