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**I need some help!****Guest**

Part 1:

Let f(x) and g(x) be polynomials.

Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8.

Suppose g(x)=0 for exactly five values of x: namely, x=-5,-3,2,4, and 8.

Is it necessarily true that g(x) is divisible by f(x)? If so, carefully explain why. If not, give an example where g(x) is not divisible by f(x).

Part 2:

Generalize: for arbitrary polynomials f(x) and g(x), what do we need to know about the zeroes (including complex zeroes) of f(x) and g(x) to infer that g(x) is divisible by f(x)?

(If your answer to Part 1 was "yes", then stating the generalization should be straightforward. If your answer to Part 1 was "no", then try to salvage the idea by imposing extra conditions as needed. Either way, prove your generalization.)

**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 79

The answer for part 1 is, by all means, yes. This is so because the condition that f(x) has the zeros -3,4 and 8 entails that:

f(x)=(x+3)×(x-4)×(x-8)

And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:

g(x)=(x+5)×(x+3)×(x-2)×(x-4)×(x-8)

and hence:

g(x)/f(x)=[(x+5)×(x+3)×(x-2)×(x-4)×(x-8)]/[(x+3)×(x-4)×(x-8)]=(x+5)×(x-2)

and since the result of the division (x+5)×(x-2) is a polynomial without a remainer, therefore g(x) is divisible by f(x). The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x). And the final statement completes the answer.

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Grantingriver wrote:

This is not quite true: the multiplicities of the zeroes are important, e.g. and .
The answer for part 1 is, by all means, yes. This is so because the condition that f(x) has the zeros -3,4 and 8 entails that:

f(x)=(x+3)×(x-4)×(x-8)

And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:

g(x)=(x+5)×(x+3)×(x-2)×(x-4)×(x-8)

and hence:

g(x)/f(x)=[(x+5)×(x+3)×(x-2)×(x-4)×(x-8)]/[(x+3)×(x-4)×(x-8)]=(x+5)×(x-2)

and since the result of the division (x+5)×(x-2) is a polynomial without a remainer, therefore g(x) is divisible by f(x). The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x). And the final statement completes the answer.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 79

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots. For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots. If there are a power for a given factor of the polynomial then there are repeted zeros and you should say something like “the polynomial f(x) has exactly the roots 4, 8 and the two duplicate roots -3”. I hope this clears your confusion.

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Grantingriver wrote:

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots.

Why? Precisely the opposite is true: if a root is repeated, then we cannot call those repeated roots 'distinct', and whether or not we count them in the same way we would distinct roots depends on the context.

Grantingriver wrote:

For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots.

I am not convinced that the word 'exactly' necessarily implies distinctness -- I would have thought that the multiplicities were worth consideration, otherwise the problem appears to be a little simplistic.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 79

Dear Zetafunc, you can alter the meaning of the word “exactly” to mean any thing you want, and it will mean that (al least to you). However, if you do not accept the meaning of “exactly” as I have stated it, then do not bother yourself about finding a prove, because the provided argument will constitute, in this situation, a prove that the supplied information is not sufficient to deduce a conclusion!! However, in this case the problem will be “awkward”, since it creates a situation that can not be proved right or wrong!! And this argument, in fact, supports the meaning which I have provided. I hope this will convince you.

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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 194
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Hi Grantigriver.

Your proof is correct, *provided* that the polynomial *f*(*x*) does not have multiple roots (i.e. all its roots have multiplicity 1). If *f*(*x*) has a multiple root, then, as zetafunc’s counterexample shows, the result no longer holds.

When I first saw this problem, I also thought like you that *f*(*x*) must be a divisor. I had forgotten all about multiplicities.

PS: IMHO the question itself is not very well worded as there is no mention of multiplicities. It would be better if we were more clear on whether there are repeated roots.

*Last edited by Alg Num Theory (2018-05-08 07:49:12)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 79

The answer for part 1 can be accomplished as follows: The condition that f(x) has the zeros -3,4 and 8 entails that:

f(x)=(x+3)ᵃ×(x-4)ᵇ×(x-8)ᶜ

And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:

g(x)=(x+5)ᵈ×(x+3)ᵉ×(x-2)ᶠ×(x-4)ᶢ×(x-8)ʰ

and hence:

g(x)/f(x)=[(x+5)ᵈ×(x+3)ᵉ×(x-2)ᶠ×(x-4)ᶢ×(x-8)ʰ]/[x+3)ᵃ×(x-4)ᵇ×(x-8)ᶜ]

so since the result of the division is (x+5)ᵈ×(x-2)ᶠ×(x+3)ᵝ×(x-4)ᵞ×(x-8)ᵟ, where β=e-a, γ=g-b and δ=h-c. Therefore, if β, γ and δ are nonnegative integers then g(x) is divisible by f(x). Otherwise it is not. The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x) and their repetetion in g(x) should be greater than or equal to their repetetion in f(x). This is the complete proof in any case.

*Last edited by Grantingriver (2018-05-08 11:04:15)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 79

The previous answer constitutes the proof of the general case (Zetafunc’s argument). However, for this problem and from its words I think the proof of the special case (the multiplicity is one for all roots) is sufficient.

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