# Proof of a corollary of fundamental theorem of algebra

1. Mar 22, 2012

### mindauggas

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

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

2. Mar 22, 2012

### Dick

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.

3. Mar 23, 2012

### 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.

4. Mar 23, 2012

### Dick

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

5. Mar 24, 2012

### mindauggas

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?

Last edited: Mar 24, 2012
6. Mar 24, 2012

### Dick

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.

7. Mar 24, 2012

### 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?

8. Mar 24, 2012

### 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".

9. Mar 25, 2012

### 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 $a(x-r_{1})$???. Lets 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: $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ ???

Last edited: Mar 25, 2012
10. Mar 25, 2012

### mindauggas

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

11. Mar 25, 2012

### 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).

12. Mar 25, 2012

### 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).

Last edited: Mar 25, 2012
13. Mar 25, 2012

### HallsofIvy

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

Last edited by a moderator: Mar 25, 2012
14. Mar 25, 2012

Thanks