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Factors and Orders of Zeros

  • Thread starter fakecop
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  • #1
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Homework Statement


Give one reason that a 4th degree polynomial function might not have four factors.


Homework Equations


(This was a test question)


The Attempt at a Solution


I wrote that a 4th degree polynomial function might contain an irreducible quadratic, such as
(x-4)(x-2)(x^2+1). Because of the factor (x^2+1), the polynomial cannot be factored over
the set of real numbers into four linear factors.

But my teacher wanted me to say that the order of the zeros may be higher than 1, so an expression such as (x-2)^2(x-4)^2 would only have two factors because there are two double roots.

My question is, doesn't the expression (x-2)^2(x-4)^2 still have four factors? The expression
(x-2)^2(x-4)^2 can be written as (x-2)(x-2)(x-4)(x-4) so it still have four factors. Just because the factors are repeated doesn't mean that it only counts as one factor, right?

The question asks about factors, not zeros. In any case, wouldn't I be right to assume that "factor" refers to linear factors?

Please help...
 

Answers and Replies

  • #2
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My question is, doesn't the expression (x-2)^2(x-4)^2 still have four factors? The expression
(x-2)^2(x-4)^2 can be written as (x-2)(x-2)(x-4)(x-4) so it still have four factors. Just because the factors are repeated doesn't mean that it only counts as one factor, right?
I agree, and I would count that as four factors as well.
In any case, wouldn't I be right to assume that "factor" refers to linear factors?
Irreducible factors I think. So (x-4)(x-2)(x^2+1) has three factors (in the real numbers).
 
  • #3
eumyang
Homework Helper
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Homework Statement


Give one reason that a 4th degree polynomial function might not have four factors.
OP: Is this the exact wording of the question? If so, I think it was badly worded. Is factoring "over the reals" supposed to be assumed?
 
  • #4
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I agree, and I would count that as four factors as well.
Irreducible factors I think. So (x-4)(x-2)(x^2+1) has three factors (in the real numbers).
Yes, my mistake. I said that you can either have four linear factors, two of which are complex conjugates, or you can have three real factors, one of which has degree 2.

OP: Is this the exact wording of the question? If so, I think it was badly worded. Is factoring "over the reals" supposed to be assumed?
The original words are: "Give one reason that a 4th degree polynomial function might not have four factors. Provide an example."


My teacher said that even though my answer was right I still can't get credit because we did not discuss factoring over the reals in class. But what's bothering me is the notion that a quartic polynomial such as (x-2)^2(x-4)^2 is considered to only have two factors. That's like saying the quadratic polynomial x^2-6x+9 only have one factor.
 
  • #5
statdad
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It seems your teacher believes only distinct (i.e. non-equal) factors should be counted separately.
 
  • #6
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Yes, my mistake. I said that you can either have four linear factors, two of which are complex conjugates, or you can have three real factors, one of which has degree 2.
Or two real factors, both with degree 2.

My teacher said that even though my answer was right I still can't get credit because we did not discuss factoring over the reals in class.
Okay... that makes the problem statement even more bizarre, as it asks about factors.
 
  • #7
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It seems your teacher believes only distinct (i.e. non-equal) factors should be counted separately.
Yes-this is the heart of the question. So what is more right mathematically speaking? Should distinct factors be counted as separate factors or not?
 
  • #8
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I also agree with you and disagree with your teacher.

Is this a high school teacher? I don't think any of my college professors would agree with this teacher, let alone not give you credit for your correct answer, even if they feel you misinterpreted the problem (and really, your "misinterpretation" makes the problem a little harder). But in high school, from what I've seen both as a student and among my colleagues later on...well, you might just have to forget about it and take solace in the fact that you're right.
 
  • #9
statdad
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yes, distinct factors are always counted as separate factors - but I think you meant to ask "should repeated factors be counted as separate factors or not?" and here the answer may not be so simple.
Consider [itex] (x-2) (x+3)^2 [/itex]
It has, when we account for multiplicity, three zeros: 2 has multiplicity 1, -3 has multiplicity 2. Note, however, there are only two distinct zeros. Technically, a similar thing is often said for factors: x-2 is a factor, (x+3) is factor of multiplicity two (One of my undergraduate profs, who was originally from Poland, referred to this as a "double factor", but I have never heard that phrasing since), giving "three" factors.

However, there are only two distinct factors, x-2 and x+3, the remaining factor being a duplicate of x+3.

Not very pleasing, I know. To agree with others: I view your response as correct, and your teacher as overly pedantic (at best).
 

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