polynomial question (1 Viewer)

[freaking_out]

the equation x^4-x^3+2x^2-2x+1 has roots a,b,c,d

i) show that none of a,b,c,d is an integer.

 

[Constip8edSkunk]

as +/- 1 are the only possible integer roots(as both the leading coefficient and the constant term are 1)sub them in and neither will get you 0, .'. there are no integer roots

 

[freaking_out]

Re: Re: polynomial question

Originally posted by Constip8edSkunk
as +/- 1 are the only possible integer roots(as both the leading coefficient and the constant term are 1)sub them in and neither will get you 0, .'. there are no integer roots

...i still don't understand, i mean, how do u know by looking at the leading coefficient and the constant term, that 1 is the only possible integer root?

 

[Constip8edSkunk]

try proving that if p/q is a root then q must be a factor of the leading coefficient and p a factor of the constant. i think it came up in the HSC b4

thats the basic theory behind it...

 

[deepulse]

Proof (for p/q thingo):

Let P(x) = ax^3 + bx^2 + cx + d

Let k be a root of P(x),

therefore, P(k) = 0

ie

ak^3 + bk^2 + ck + d = 0

d = -(ak^3 + bk^2 + ck)

d = -k(ak^2 + bk + c)

And since a,b,c,d are all integers (even if they're not, just multiply them out...u get the idea), d is divisible by k

Then u would do the same, but 1/k, then ull get

a + bk + ck^2 + dk^3 = 0 (by multipying by k^3)

a = -k(dk^2 + ck + b)

Thats one way atleast

 

[spice girl]

they call it the rational root theorem

if x = p/q is a rational root, then p divides the constant term, and q divides the leading coefficient.

 

[underthesun]

is that quoteable in the HSC?

 

[freaking_out]

Originally posted by Constip8edSkunk
try proving that if p/q is a root then q must be a factor of the leading coefficient and p a factor of the constant. i think it came up in the HSC b4

thats the basic theory behind it...

this question is from a catholic trial paper (which hardly reflects the hsc)....

 

[Newbie]

Originally posted by freaking_out
this question is from a catholic trial paper (which hardly reflects the hsc)....

i've seen a harder variant of this in a past hsc paper actually
forgot which one though

it was a q7 i think

 

[ND]

Originally posted by Newbie
i've seen a harder variant of this in a past hsc paper actually
forgot which one though

it was a q7 i think

The '01 paper had a q7 with the first part similar to that.

 

[freaking_out]

Originally posted by ND
The '01 paper had a q7 with the first part similar to that.

yeah i saw it now, but i wanna know how r we meant to know that...like is it a standard proof that we meant to know...if so can someone tell me where in any textbook can i find this theorem explained?

 

[ND]

yeah i saw it now, but i wanna know how r we meant to know that...like is it a standard proof that we meant to know...if so can someone tell me where in any textbook can i find this theorem explained?

You're supposed to be able to prove anything asked in a 4u paper. You shouldn't be prepareing for 4u exams by memorising proofs and stuff, that is pointless, you should learn to prove stuff. I'll try and explain this theorem to you:

p/q is a roots of P(x), where p and q are integers with no common factor.

P(x) = ax^3+bx^2+cx+d
P(p/q)= a(p/q)^3 + b(p/q)^2 + c(p/q) + d = 0
ap^3 + bp^2q + cpq^2 + dq^3 = 0
ap^3/q = -(bp^2+cpq+dq^2)
Now the RHS is an integer, so the LHS must also be an integer, and because p/q isn't an integer (from the definition), then a/q must be.
.'. q divides a.

To prove p divides d do the same thing only with dq^3/p.

 

[Saul]

if you're asked to prove it though, you should do it for general polynomial... same proof tho.

 

[ND]

Originally posted by Saul
if you're asked to prove it though, you should do it for general polynomial... same proof tho.

Yeh but usually they'll give you the polynomial anyway, so you just do it with that.

 

[freaking_out]

Originally posted by ND
You're supposed to be able to prove anything asked in a 4u paper. You shouldn't be prepareing for 4u exams by memorising proofs and stuff, that is pointless, you should learn to prove stuff.

yeah i know, but spice girl was saying that it was a theorem, so thats why i asked since i've never seen anythin' like this, and also it was assumed that i knew it to do the question above ....thanx for the explanation though, now i understand

 

[Hotdog1]

I would like to reiterate:

Originally posted by underthesun
is that quoteable in the HSC?

Can we just say it is the rational root theorem without proof?