Understanding Zeros Theorem Proof  TOPIC_SOLVED

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Understanding Zeros Theorem Proof  TOPIC_SOLVED

Postby jwroblewski44 on Sat Dec 21, 2013 3:00 am

I am having trouble understanding the proof in my text for the Zeros Theorem. It goes like this:

"Proof: Let P be a polynomial of degree n. By the complete factorization theorem
P(x) = a(x-c1)(x-c2)(x-cn) //the numbers and letter n and i are subscript
Now suppose that c is a zero of P other than c1,c2,.....cn. Then
P(c) = a(c-c1)(c-c2)(c-cn) = 0
Thus, by the Zero-Product Property, one of the factors c-ci, must be 0, so c=ci for some i. It follows that P has exactly the n zeros c1,c2,.....cn."

I was understanding up until they start talking about c-ci must be 0 so c=ci for some i. I'm not sure what they are saying.

Can someone help explain this to me? Thank you in advance!
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Postby stapel_eliz on Sat Dec 21, 2013 12:58 pm

jwroblewski44 wrote:
Proof: Let P be a polynomial of degree n. By the complete factorization theorem, P(x) = a(x - c1)(x - c2) ... (x - cn). Now suppose that c is a zero of P other than c1, c2, ..., cn). Then P(c) = a(c - c1)(c - c2) ... (c - cn) = 0

Thus, by the Zero-Product Property, one of the factors c - ci, must be 0, so c = ci for some i. It follows that P has exactly the n zeros c1, c2, ..., cn.

I was understanding up until they start talking about "c - ci must be 0 so c = ci for some i." I'm not sure what they are saying. Can someone help explain this to me?

To be a "zero" of the polynomial, the x-value ("c", in the case) must make the polynomial equal to zero. So they've plugged "c" in for "x" in the factored form of the polynomial. By definition then, the polynomial is zero with "c" in for "x".

The polynomial is in factored form. If none of the factors is equal to zero, then the polynomial, being the product of those non-zero values, would also evaluate to some non-zero value. But the polynomial is (by definition of "zero") equal to zero, so at least one of the factors of the polynomial must be zero.

Pick one of those "equal to zero" factors. Since they've plugged "c" in for "x" to get this "equal to zero" factor, what does that factor look like? Setting that factor equal to zero (in other words, creating an equation which reflects what we know the value of the factor to be), what is the resulting value of "c"? :wink:
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Re: Understanding Zeros Theorem Proof

Postby jwroblewski44 on Sat Dec 21, 2013 3:36 pm

So when they say 'ci' they aren't using i to mean sqrt(-1)? The use of 'i' has really confused me.

EDIT: A question I forgot to mention in my OP, how is there a zero, c, that is not c1,c2....cn?

I understand that every polynomial has a value that will result in the polynomial equaling zero. That was covered a lot earlier in the book. But I still can't see how this proof shows us how every polynomial has exactly as many factors as the degree of the polynomial?

PS Thanks for taking time to help a stranger!
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Postby stapel_eliz on Sun Dec 22, 2013 1:42 am

jwroblewski44 wrote:So when they say 'ci' they aren't using i to mean sqrt(-1)? The use of 'i' has really confused me.

No; all the subscripts are just that: counters which have been subscripted.

jwroblewski44 wrote:EDIT: A question I forgot to mention in my OP, how is there a zero, c, that is not c1,c2....cn?

The proof says exactly that: any zero "c" must be one of the "n" zeroes denoted "ci" for i = 1 through i = n.

jwroblewski44 wrote:I still can't see how this proof shows us how every polynomial has exactly as many factors as the degree of the polynomial?

The proof starts with a zero, "c", and then shows that this zero must be equal to one of the n zeroes, ci. Since any zero c must be one of the n zeroes denoted in the factored form, then the polynomial can't have any more than n zeroes. Since the polynomial has n factors and thus n corresponding zeroes, then the polynomial has at least n zeroes. Since the polynomial has "at least" n zeroes but "no more than" n zeroes, then the polynomial must have exactly n zeroes. :wink:
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Re: Understanding Zeros Theorem Proof

Postby jwroblewski44 on Sun Dec 29, 2013 8:50 pm

Sorry for the large time span between posts. That helped clear it up! Thanks again.
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