Question

Prove:

Let V(x) and D(x) be polynomials,
then there exists exactly one polynomial Q(x) and exactly one polynomial R(x) where:
V(x) = Q(x)*D(x)+R(x)
and
(R(x)=0 or degree[R(x)] 13 years ago

Answer 1

rewrite like V(x)/Q(x) = D(x) + R(x)/Q(x)

Answer 2

yes, but then? :D

Answer 3

*

Answer 4

hmm, not sure. I though we were going to be able to use the remainder theorem in a pretty simple way to show this, but it's not coming to me...
I'll have to think about it some more, maybe someone can jump in if they have an idea..

Answer 5

Well, it is quite possible related to the remainder theorem. It is stated in my textbook, just above the remainder theorem. But no proof is given so I find it hard to understand.

Answer 6

numbers are called polynomials of degree 0, so it seems to be a more general approach to the thrm

Answer 7

Unless the number is 0. Then it is a polynomial of degree "undefined" or degree "-infinity"

Answer 8

correct

Answer 9

think of polynomials as encrypted "numbers".

Answer 10

since they are functions, then for any value of x, there is a distinct value for y that results

Answer 11

I guess 12/3 = 4 and R=0. degree of R(x) has to be less than degree of D(x) or R(x) has to be zero. So I guess that's a good start, since you can't give any other Q(x) than 4 where R=0

Answer 12

But it's not a proof...

Answer 13

the proof would start out like this most likely

let\[P(x) = p_nx^n+p_{n-1}x^{n-1}+p_{n-2}x^{n-2}+...+p_2x^2+p_1x+p_0\]let\[Q(x) = q_nx^n+q_{n-1}x^{n-1}+q_{n-2}x^{n-2}+...+q_2x^2+q_1x+q_0\]

Answer 14

There is one polynomial, D(x) such that\[D(x)=\frac{p_nx^n}{q_nx^n}+...\]
maybe swap out ns with as and bs to indicate the possibility of different degrees

Answer 15

\[q_2x^2+q_1x+q_0~|~p_3x^3+p_2x^2+p_1x+p_0\]


D(x) =\(\large \frac{p_3x^3}{q_2x^2}+\frac{}{}+\frac{}{}+\)
---------------------------
q_2x^2+q_1x+q_0 | \(p_3x^3+p_2x^2+p_1x+p_0\)
\(-p_3x^3\)

latex needs a long hand division mode ....

Answer 16

also search about Division Algorithm of polynomials

Answer 17

But can you divide a polynomial of lower degree by one of higher degree? Wouldn't that disprove the theorem?

Answer 18

guys, why are u complicating things ., showing the uniqueness is real simple.
use contradiction, and prove that its not possible to have 2 r(x)'s.

Answer 19

you cant divide a lower degree by a higher and still get a polynomial back

Answer 20

or something like that

Answer 21

Yes, doesn't that disprove the theorem?