Polynomials problem help

  • Thread starter Ted123
  • Start date
  • #1
446
0

Homework Statement



[PLAIN]http://img443.imageshack.us/img443/3096/questiond.jpg [Broken]

The Attempt at a Solution



If [itex]P(z)Q(z)=0[/itex] then

[itex]\displaystyle a_0b_0 + (a_0b_1 + a_1 b_0)z + ... + \left( \sum_{i=0}^k a_i b_{k-i} \right) z^k + ... + a_n b_m z^{n+m} =0[/itex]

Now what? Equate coefficients?

[itex]a_0 b_0 =0 \Rightarrow a_0 = 0 \; \text{or}\; b_0 = 0[/itex]

[itex]a_0 = 0 \Rightarrow a_1 b_0 =0 \Rightarrow a_1 = 0 \; \text{or}\; b_0 =0[/itex]

[itex]b_0 =0 \Rightarrow a_0 b_1 = 0 \Rightarrow b_1 = 0 \; \text{or}\; a_0 = 0[/itex]

...
 
Last edited by a moderator:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Ted123! :smile:

Hint: start at zn+m. :wink:
 
  • #3
446
0


Hi Ted123! :smile:

Hint: start at zn+m. :wink:

Doing this gives

[itex]a_n = 0\;\text{or}\; b_n = 0[/itex] but what next?

Would it be better to use the fact that:

If P isn't identically zero, it has at most n roots.
If Q isn't identically zero, it has at most m roots.

So, if neither P and Q are identically zero, PQ has at most m+n roots. How could I use this?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
… but what next?

think! :smile:

(how are an and bm defined?)
 
  • #5
446
0


think! :smile:

(how are an and bm defined?)

Well in the question before this it states that [itex]a_n \neq 0[/itex] and [itex]b_m \neq 0[/itex] and it says in this question let P and Q be complex polynomials as in the previous question (I missed these conditions out when I copied the polynomials from the previous question).

So if [itex]a_n b_m =0[/itex] but [itex]a_n , b_m \neq 0[/itex] what do we have?
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
a contradiction!!

soooo … ? :smile:
 
  • #7
446
0


a contradiction!!

soooo … ? :smile:

[itex]P(z)Q(z) \neq 0[/itex]
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,832
251
Yes … you've proved that if P and Q are not zero, then PQ is not zero. :smile:
 
  • #9
446
0


Yes … you've proved that if P and Q are not zero, then PQ is not zero. :smile:

So does this prove the converse (which I need to prove) that if PQ=0 then P or Q are 0? Oh yes - I see it does now :smile:
 
Last edited:

Related Threads on Polynomials problem help

  • Last Post
Replies
1
Views
884
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
5
Views
991
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
5
Views
4K
Top