Proof of a corollary of fundamental theorem of algebra

[mindauggas]

Homework Statement

Assuming the validity of the fundamental theorem of algebra, prove the corollary that:

Every polynomial of positive degree n has a factorization of the form:

P(x)=a_{n}(x-r_{1})...(x-r_{n}) where r_{i} aren't necessarily distinct.

Homework Equations

Fundamental Theorem of Algebra: Every polynomial of positive degree with complex coefficients has at least one complex zero.

The Attempt at a Solution

Does this even require proof? Don't know where to begin...

 

[Dick]

Homework Statement

Assuming the validity of the fundamental theorem of algebra, prove the corollary that:

Every polynomial of positive degree n has a factorization of the form:

P(x)=a_{n}(x-r_{1})...(x-r_{n}) where r_{i} aren't necessarily distinct.

Homework Equations

Fundamental Theorem of Algebra: Every polynomial of positive degree with complex coefficients has at least one complex zero.

The Attempt at a Solution

Does this even require proof? Don't know where to begin...

Sure it requires proof. Start by saying P(x) is has a root r1. Show (x-r1) divides P(x), so you can write P(x)=(x-r1)P1(x) where P1(x) has degree n-1. Now apply the same thing to P1(x). Etc. If you want to be more formal you prove it by induction.

 

[mindauggas]

Attempt:

(1) We have a polynomial P(x)=a(x-r_{1})

(2) x-r_{1} is a factor if and only if r_{1} is a zero (Remainder theorem)

(3) P(r_{1})=a(r_{1}-r_{1}) which is zero, therefore r_{1} is a zero and x-r_{1} is a factor.

At this point I have a question: by what theorem can now take the step: (4) P(x)=a(x-r_{1})(x-r_{n-1}) or should this be obvious? Because I don't understand why is this the case.

 

[Dick]

P(x) is a polynomial of degree n. You've only got a polynomial of degree 1. Reread my last post.

 

[mindauggas]

P(x) is a polynomial of degree n. You've only got a polynomial of degree 1. Reread my last post.

Start by saying P(x) is has a root r1

But I can't assume that P(x) is a polynomial o degree n and than just write: P(x)=a(x-r_{1}) where do I indicate the degree? Should I write: P(x)=a(x-r_{1})^{n} ? Or something?

 

[Dick]

But I can't assume that P(x) is a polynomial o degree n and than just write: P(x)=a(x-r_{1}) where do I indicate the degree? Should I write: P(x)=a(x-r_{1})^{n} ? Or something?

I would write it as P(x)=(x-r_{1})P_1(x). When you divide x-r_1 into P(x) you are going to get another polynomial, not just a constant.

 

[mindauggas]

Attempt #2:

(1) We denote a polynomial of degree n as P(x)=a(x-r_{1})P_{1}(x)

(2) x-r_{1} is a factor if and only if r_{1} is a zero (Remainder theorem)

(3) P(r_{1})=a(r_{1}-r_{1})P_{1}(r_{1}) which is (NOT?)zero, therefore r_{1} is (NOT?)a zero and x-r_{1} is (NOT?) a factor.

It doesn't divide then? Or have I misunderstood smth?

 

[HallsofIvy]

I would be inclined to do this as a "proof by induction" on n, the degree of the polynomial. You have done the "n= 1" case. Now show that "if a polynomial of degree k can be factored as linear factors, so can a polynomial of degree k+1".

 

[mindauggas]

(1) We have a statement P_{1} "a polynomial a(x-r_{1}) has expresion x-r_{1} as a factor", which is equivalent to saying that it can be factored as a(x-r_{1})?. Let's call the first polynomial P_{1}(x) in accordance to first statement and denote P_{n}(x) a polynomial corresponding to the statement P_{n} for the polynomial a(x-r_{1})(x-r_{2})...(x-r_{n})

(2) Now, x-r_{1} is a factor if and only if r_{1} is a zero (Remainder theorem)

(3) Since P_{1}(r_{1})=a(r_{1}-r_{1}) is equal to zero, therefore r_{1} is a zero and x-r_{1} is a factor of the polynomial.

(4) Assume that statement P_{k} is true.

(5) P_{k+1} is the statement: a polynomial a(x-r_{1})(x-r_{2})...(x-r_{k})(x-r_{k+1}) has expresion x-r_{k+1} as a factor. Repeating (2) and (3) with r_{k+1} we get P_{k+1}(k+1)=0

(6)Therefore every polynomial of positive degree n has a factorization of the form:

P(x)=a_{n}(x-r_{1})...(x-r_{n}) where r_{i} aren't necessarily distinct.

Q.E.D. ? I don't think so.

Shouldn't I start with general rule for a polynomial: f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0} ?

 

[mindauggas]

Can someone approve that this constitutes a proof of what I intended to prove? Maybe additional qualifications are needed - some steps are missing?

 

[Dick]

If you are doing induction P_k should be the statement "any polynomial of degree k can be expressed in the form a (x-r_1) (x-r_2) ... (x-r_k)".

You can do the k=1 case without using your theorem at all. f(x)=a_1 x + a_0 is a polynomial of degree 1. You can write that as f(x)=a_1 (x + \frac{a_0}{a_1}) Since your root is r_1=(-a_0/a_1) that expresses f(x) in the form a(x-r). To prove P_k implies P_{k+1}, pick a polynomial p(x) of degree k+1. The fundamental theorem of algebra says p(x) has a root, call it r_{k+1}. So p(x)=(x-r_{k+1})*q(x) where q(x) has degree k. Now use your induction hypothesis on q(x).

 

[mindauggas]

(1) Assume the statement P_{k}: "the polynomial q(x) od degree k can be written as q(x)=a_{k}(x-r_{1})...(x-r_{k})" is true.

(2) Now the statement P_{k+1} is about the polynomial p(x)=(x-r_{k+1})q(x) which can be rewritten as: p(x)=(x-r_{k+1})a_{k}(x-r_{1})...(x-r_{k}).

(3) Since we've assumed P_{k} is true, and P_{k+1} was shown to be true previously (I didn't write the statement, but its trivial) - QED (?)

Can anyone give feedback, especially for the step (2).

 

[HallsofIvy]

You need to state explicitely, and support,
"If P_k+1 is a polynomial of degree k+1 and P_k+1(x)= (x- r)Q(x), then Q(x) is a polynomial of degree k".

 

[mindauggas]

Thanks