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Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 1:27 pm UTC
by Paradoxica
P(x) = axn+1 + bxn + r
P(x) has a double root of x = 1
show that a = n, and b = -(n+1)
What I have done so far:
P(x) has a double root of x = 1, so P'(x) has a root of x = 1
P'(x) = (n+1)axn + nbxn-1
P'(1) = (n+1)a + nb = 0
(n+1)a = -nb
n ≠ n+1
So it follows that a = -n and b = n+1 or a = n and b = -(n+1)
I don't know where to go from here, and how I can eliminate the other case.
Either I'm stupid or this question is wrong.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 2:11 pm UTC
by gmalivuk
Paradoxica wrote:P(x) = axn+1 + bxn + r
P(x) has a double root of x = 1
show that a = n, and b = -(n+1)
What I have done so far:
P(x) has a double root of x = 1, so P'(x) has a root of x = 1
P'(x) = (n+1)axn + nbxn-1
P'(1) = (n+1)a + nb = 0
(n+1)a = -nb
n ≠ n+1
So it follows that a = -n and b = n+1 or a = n and b = -(n+1)
I don't know where to go from here, and how I can eliminate the other case.
Either I'm stupid or this question is wrong.

You've used the root of P'(x) to get some information about a and b.

You have not yet used the root of P(x) to get any additional information.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 2:20 pm UTC
by Paradoxica
gmalivuk wrote:You've used the root of P'(x) to get some information about a and b.

You have not yet used the root of P(x) to get any additional information.


I'm still not getting anywhere. Substitution, isolation, solving simultaneously. I've tried all of those.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 6:51 pm UTC
by gmalivuk
You shouldn't need to do anything super complicated. You know there's a root at x=1, so P(1)=0. What does that tell you about a and b?

Also, it's a double root, which presumably means it's not a triple root. What does that tell you about P''(x)? (I don't know if that's necessary, but it might be enough to distinguish between one result and its negative.)

Edit: The equation (n+1)a = -nb does NOT tell you a = +/- n and b = - /+(n+1)
All it tells you is that a/b = - n/(n+1). Even if you knew the signs of a and b, what you have so far wouldn't prove what you're looking for.

(Actually, it looks like there's still some other missing information, because r is another unknown which, if it can freely vary, doesn't give us any new information when we look at P(1)=0.)

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 7:01 pm UTC
by Nicias
As stated, the problem is incomplete.

If (a,b,r) is a solution so is (k a, kb, kr).

Maybe you need to assume that r=1?

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 8:46 pm UTC
by Paradoxica
Nicias wrote:As stated, the problem is incomplete.

If (a,b,r) is a solution so is (k a, kb, kr).

Maybe you need to assume that r=1?


P(1) = a + b + r =0 (1)
P'(1)= (n+1)a + nb = 0 (2)
n+1 ≠ n
|(n+1)a| = |nb|
n+1|a| = n|b|
n and n+1 are positive integers, although I'm not sure how to use this fact.
|a| = n
|b| = n+1
If a and b were the same value, this would imply n = n+1, an impossibility.
Also, since n and n+1 cannot be both even or both odd, then both terms in (1) are even, in order to cancel.
Another result is that if a is negative, b is positive and vice versa, since they cancel.
This generates two cases.
Case 1: a = -n, b = n+1
Case 2: a = n, n = -(n+1)
Note that a + b + r = 0
Applying this fact to case 1 yields r= -1
Similarly, case 2 yields r = 1
So P(x) = axn+1 + bxn ±1
I don't see how (ka,kb,kr) would be a solution.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 21, 2015 8:55 pm UTC
by Nicias
Paradoxica wrote:
Nicias wrote:As stated, the problem is incomplete.

If (a,b,r) is a solution so is (k a, kb, kr).

Maybe you need to assume that r=1?


|(n+1)a| = |nb|
n+1|a| = n|b|
n and n+1 are positive integers, although I'm not sure how to use this fact.
|a| = n
|b| = n+1

This is your problem. This does not follow.
(n+1)a =-n(b) does not imply that a=-n and b=n+1 you could also have a=-2n, b=2(n+1), or a=107n b=-107(n+1).

So for example, if n=5. Then P(x) = 5x^6-6x^5+1. P(1) = 0, and P'(1)=0.
But P(x) = 10x^6-12x^5+2 works too. P(1) = 10-12+2=0, P'(1) = 60-60=0.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sun Feb 22, 2015 3:56 am UTC
by PM 2Ring
Are a and b restricted to integers?

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sun Feb 22, 2015 12:08 pm UTC
by Paradoxica
PM 2Ring wrote:Are a and b restricted to integers?

Well according to the post made to correct my assumption, no.
ka, kb, kr is a solution to the system of equations, and k can be any real number, so therefore, a and b do not have to be integers.
Although the answer to my homework is asking to show that they are. :/
Unfortunately, no value of r is given, and r depends on a and b. Three variables, two equations. I'm going to go on a whim and assume the question is incorrectly stated.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sun Feb 22, 2015 1:51 pm UTC
by gmalivuk
Paradoxica wrote:
PM 2Ring wrote:Are a and b restricted to integers?

Well according to the post made to correct my assumption, no.
ka, kb, kr is a solution to the system of equations, and k can be any real number, so therefore, a and b do not have to be integers.
No, "restricted to integers' means the problem itself tells you they're integers.

It still doesn't help, though, because if there's one integer solution there are infinitely many.

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Fri Feb 27, 2015 12:03 pm UTC
by Paradoxica
Turns out the question was incorrectly stated. r = -1

Re: Polynomials : Multiplicity of roots Homework problem

Posted: Sat Feb 28, 2015 9:21 am UTC
by flownt
Paradoxica wrote:Turns out the question was incorrectly stated. r = -1

Presumably you now managed to find the answer?