Rational roots of a polynomial (2 Viewers)

[vds700]

I was doing thiis question from the independent 2008 trial

5a) P(x) = x^4 - 4x^3 + 5x^2 -2x -2

(ii) Explain why the real roots cannot be rational.

In the solution they say "Any rational roots of P(x) = 0 must be factors of 2". They sub in factors of 2 and none of them are roots, so there are no rational roots.

wtf I have never heard of this before. Looking through Cambridge and Fitzpatrick, I cannot locate any such rule.

Can someone please explain and provide the general rule for rational roots of polynomials. Thanks

 

[lyounamu]

I was doing thiis question from the independent 2008 trial

5a) P(x) = x^4 - 4x^3 + 5x^2 -2x -2

(ii) Explain why the real roots cannot be rational.

In the solution they say "Any rational roots of P(x) = 0 must be factors of 2". They sub in factors of 2 and none of them are roots, so there are no rational roots.

wtf I have never heard of this before. Looking through Cambridge and Fitzpatrick, I cannot locate any such rule.

Can someone please explain and provide the general rule for rational roots of polynomials. Thanks

I don't understand their solution too. But personally I would use the polynomials approximation method to say that the root cannot be rational.

 

[undalay]

LOL.

consider (x+a)(x+b)(x+c)(x+d) = x^4 - 4x^3 + 5x^2 -2x -2

equating constants abcd = -2

So the four roots a,b,c,d must multiply to equal -2

NOW if they were rational, the only numbers capable of doing so are 1,-1,2,-2!

basically u sub those four nubmers in the equation, and show they dont equal 0!

SO SINCE 1,-1,2,-2 arent roots and they are the only possible rational roots. IT HAS NO RATIONAL ROOTS : O : O

edit: u can skip 1,-1 if u want. Furthermore it cant be fractions cause all coefficients are whole numbers and leading coefficient is 1.

edit edit:


For polynomial ax^n + bx^n-1 + ... + Z

the rational roots must be positive negative factors of z/a

 

[lolokay]

I don't understand their solution too. But personally I would use the polynomials approximation method to say that the root cannot be rational.

lol that seems to be your answer for everything (well, a couple things at least)

the monic + integer coefficients :. rational roots are integers thing is pretty cool. what's the proof for it?

 

[vds700]

LOL.

consider (x+a)(x+b)(x+c)(x+d) = x^4 - 4x^3 + 5x^2 -2x -2

equating constants abcd = -2

So the four roots a,b,c,d must multiply to equal -2

NOW if they were rational, the only numbers capable of doing so are 1,-1,2,-2!

basically u sub those four nubmers in the equation, and show they dont equal 0!

SO SINCE 1,-1,2,-2 arent roots and they are the only possible rational roots. IT HAS NO RATIONAL ROOTS : O : O

edit: u can skip 1,-1 if u want. Furthermore it cant be fractions cause all coefficients are whole numbers and leading coefficient is 1.

edit edit:


For polynomial ax^n + bx^n-1 + ... + Z

the rational roots must be positive negative factors of z/a

Thanks mate, makes sense now

 

[shaon0]

I was doing thiis question from the independent 2008 trial

5a) P(x) = x^4 - 4x^3 + 5x^2 -2x -2

(ii) Explain why the real roots cannot be rational.

In the solution they say "Any rational roots of P(x) = 0 must be factors of 2". They sub in factors of 2 and none of them are roots, so there are no rational roots.

wtf I have never heard of this before. Looking through Cambridge and Fitzpatrick, I cannot locate any such rule.

Can someone please explain and provide the general rule for rational roots of polynomials. Thanks

lol, in our yearly exams we had a 2 mark question saying something similar to that.

 

[lyounamu]

lol that seems to be your answer for everything (well, a couple things at least)

the monic + integer coefficients :. rational roots are integers thing is pretty cool. what's the proof for it?

yeah, but that's like the easiest method, lol

 

[Iruka]

Apparently it is called the Rational Root Theorem.

http://en.wikipedia.org/wiki/Rational_root_theorem

 

[AMorris]

undalay's proof doesn't work. What about if the roots were -1/2, 1/2, 1/2, 16? they are all rational but none of them are the ones you suggested.

Of course the theorem still holds but its proof is somewhat different.

First, we suppose that x = p/q (p, q are integers with no common factors) was a rational root of the polynomial, P(x). Then we know taht P(p/q) = 0.

expanding this out tells us that:

a_n*p^n/q^n + ... + a_0 = 0

multiplying this by q^n:

a_n*p^n + a_n*p^(n-1)*q + ... + a_0*q^n = 0

Now p divides the right hand side (p*0 = 0) and p divides every term but the last on the left hand side. Hence, p must divide the last term (modular arithmetic would formalise this). since p doesn't divide q, p must then divide a_0. by a similar argument for q, we get that q divides a_n.

So returning to the specific question, this means that any rational root of the form x=p/q must be such that p divides -2 and q divides 1. so the options are p = 1, -1, 2, -2 and q =1, -1. so x = 1, 2, -1, -2 (the options that the solution tells us).

 

[shaon0]

undalay's proof doesn't work. What about if the roots were -1/2, 1/2, 1/2, 16? they are all rational but none of them are the ones you suggested.

Of course the theorem still holds but its proof is somewhat different.

First, we suppose that x = p/q (p, q are integers with no common factors) was a rational root of the polynomial, P(x). Then we know taht P(p/q) = 0.

expanding this out tells us that:

a_n*p^n/q^n + ... + a_0 = 0

multiplying this by q^n:

a_n*p^n + a_n*p^(n-1)*q + ... + a_0*q^n = 0

Now p divides the right hand side (p*0 = 0) and p divides every term but the last on the left hand side. Hence, p must divide the last term (modular arithmetic would formalise this). since p doesn't divide q, p must then divide a_0. by a similar argument for q, we get that q divides a_n.

So returning to the specific question, this means that any rational root of the form x=p/q must be such that p divides -2 and q divides 1. so the options are p = 1, -1, 2, -2 and q =1, -1. so x = 1, 2, -1, -2 (the options that the solution tells us).

Would we be required to produce exactly that in a test given the same question?

 

[pLuvia]

Would we be required to produce exactly that in a test given the same question?

There have been some questions in the HSC exams which use the method of letting x=p/q and is rational and you have to use contradiction to prove something didn't work. Not sure which paper it was but it was a question like prove pi is irrational.

They may stick something like this in future exams.

 

[shaon0]

There have been some questions in the HSC exams which use the method of letting x=p/q and is rational and you have to use contradiction to prove something didn't work. Not sure which paper it was but it was a question like prove pi is irrational.

They may stick something like this in future exams.

Oh alrite. Can i use that theorem then even if its not in the syllabus?

 

[undalay]

undalay's proof doesn't work. What about if the roots were -1/2, 1/2, 1/2, 16? they are all rational but none of them are the ones you suggested.

Of course the theorem still holds but its proof is somewhat different.

First, we suppose that x = p/q (p, q are integers with no common factors) was a rational root of the polynomial, P(x). Then we know taht P(p/q) = 0.

expanding this out tells us that:

a_n*p^n/q^n + ... + a_0 = 0

multiplying this by q^n:

a_n*p^n + a_n*p^(n-1)*q + ... + a_0*q^n = 0

Now p divides the right hand side (p*0 = 0) and p divides every term but the last on the left hand side. Hence, p must divide the last term (modular arithmetic would formalise this). since p doesn't divide q, p must then divide a_0. by a similar argument for q, we get that q divides a_n.

So returning to the specific question, this means that any rational root of the form x=p/q must be such that p divides -2 and q divides 1. so the options are p = 1, -1, 2, -2 and q =1, -1. so x = 1, 2, -1, -2 (the options that the solution tells us).

obviously cant b fractions brah

 

[pLuvia]

I never knew it was a theorem because it was a solution for one of the HSC papers

It's more using the method of contradiction to solve a problem, basically you assume that the thing takes this form but later proving that it doesn't work hence that thing can't take that form by contradiction

 

[tommykins]

It was in 2003 I think, last question where it says prove e is irrational.

 

[midifile]

obviously cant b fractions brah

yes it can... fractions can be rational too.

amorris's proof is correct (although it isnt written out in full)

my teacher taught our class that proof

 

[shaon0]

It was in 2003 I think, last question where it says prove e is irrational.

How long is the proof of e is irrational meant to take? I saw a proof on the internet which takes a few lines but it isn't in the HSC syllabus.
Are you still allowed to use a prooof even if its not in the syllabus?

 

[Js^-1]

Hmm, I think you can use anything, so long as it's mathematically viable and logically consistent, unless it says "hence..."
If you do use something out of the syllabus though, you can't necessarily quote it as fact. You may need to show where it came from, and maybe prove it. Quoting a formula and subbing it in won't be enough. (That's just in general).

 

[undalay]

yes it can... fractions can be rational too.

amorris's proof is correct (although it isnt written out in full)

my teacher taught our class that proof

no im saying for that equation is cant be fractions (show me a monic polynomial with integer coefficients that has fraction roots) .
Obviously fractions are rational.

Any my post wasn't meant to be a proof, rather its suppose to stimulate thoughts/intuition.
proofs are lameee.

 

[shaon0]

Hmm, I think you can use anything, so long as it's mathematically viable and logically consistent, unless it says "hence..."
If you do use something out of the syllabus though, you can't necessarily quote it as fact. You may need to show where it came from, and maybe prove it. Quoting a formula and subbing it in won't be enough. (That's just in general).

So i would have to prove it on a HSC paper. Ok thanks.