Question

im I correct? are these all considered polynomial?
2x^2-15x+10

8x-2^-2

9x^-3+8

Answer 1

HINT: Do they have a degree equal to or more than \(1\)?

Answer 2

And also, positive integers?

Answer 3

Or I mean, the lowest power greater than or equal to 0.

Answer 4

yes they all do ?

Answer 5

So much words... polynomials can't have unknowns in the numerator with negative exponents...

Answer 6

So, butterfly... some of these aren't quite polynomials ;)
Have you pinpointed them?

Answer 7

terenz im confused lol i think this is the only one that is 2x^2-15x+10

Answer 8

(Y)

Answer 9

Well, stay your confusion, because you're RIGHT :D
The second one has a negative exponent
So does the third.

Answer 10

:')

Answer 11

NOTE: Just because all exponents are positive doesn't guarantee polynomial-ness ;)
Case in point...
\[\huge x^\frac52+x^2+x+1\]Is not a polynomial.
Polynomials must have positive integers as exponents (in the numerator, of course)

Answer 12

and these are not polynomials because they have division correct ?
2m^4-r^3/5

5x^2y^3/z

Answer 13

@ParthKohli
Constants are polynomials. So degree may be zero ;)

Answer 14

Yeah, see the third post. The first and second were wrong :-)

Answer 15

@ParthKohli TL;DR
LOL Just kidding, I only saw that now :D
Sorry, my bad

@mariposa007
Well, when division is involved, it MAY still be a polynomial, provided there are no unknowns (variables) in the denominator that are raised to a positive exponent.

Answer 16

Here's how think about it.\[\large x^{\frac{5}{2}} = x^{5 \times 2^{-1}}\]Negative power!

But again.\[\large x^2 = x^{-2 \times -1}\]

Answer 17

OK, just follow Terence. I don't wanna confuse you =_=

Answer 18

@ParthKohli
And I thought I had the tendency to complicate things..

Hey, Miss Butterfly... @mariposa007 you have an idea now?

Answer 19

LOL

Answer 20

@terenzreignz I love you, man.

Answer 21

Nooooooo @ajprincess. Not me!

Answer 22

I know. I love me too :)
LOL
You know, @ParthKohli I really don't know how to respond to that, so take this alpha...
\[\huge \alpha\]

Answer 23

\[\Huge \alpha\]I have a bigger alpha for you. :-)

Answer 24

I do love alphas...

Answer 25

guys guys focus LOL you have lost me. i know you might think is simple but if i am correct then 2m^4-r^3/5

Answer 26

That's a polynomial.

Answer 27

is a poly/ correct?

Answer 28

Yupppp.

Answer 29

Yes. Precisely, that is still a polynomial, because it's basically
\[\huge \frac25m^4-\frac15r^3\]

Answer 30

Nice reasoning, Miss Butterfly :D

Answer 31

It belongs to the polynomial ring \(\mathbb{Z}[m]\).

Answer 32

:3

Answer 33

but not this one . this is not a poly 5x^2y^3/z

Answer 34

Correct-o!

Answer 35

@ParthKohli What are you doing?
@ParthKohli STAHP

@mariposa007 Yeah, because that is basically
\[\huge 5x^2y^3z^{-1}\]

Answer 36

;-P

Answer 37

And you have a negative exponent...

Answer 38

Hehe, I was having fun. :-)

Answer 39

you guys have been awesome!!!

Answer 40

Yeah. You too, Miss Butterfly :P
The name's Terence ;)

Answer 41

Thanks Terence!!!

Answer 42

Say thanks to the sidekick too.

Answer 43

:-)

Answer 44

She already did :P
Have a nice day :D

Answer 45

@ParthKohli Thanks LOL you both have been awesome how can i follow you guys ? i am new to this site

Answer 46

You can become a fan of Terence.

Answer 47

Follow me? You mean like a stalker? Dont' you dare -.-
LOL just kidding :)
Just enjoy this site :D

@ParthKohli You're so sweet :)

Answer 48

LOL !!!

Answer 49

@terenzreignz I love you too. :-)

Answer 50

I'm all out of alphas, take this tau, instead...
\[\Huge \tau\]

Answer 51

<3

Answer 52

I give you this lambda.\[\Huge \lambda\]

Answer 53

Wait a sec... what do you mean by
\[\large \mathbb{Z}[m]\]@ParthKohli
For one thing, m isn't the only unknown, and even if it is, the coefficients aren't integers
ಠ.ಠ

LOL

by the way, @mariposa007 ignore this ;)

Answer 54

OMG, wait lol, I thought the \(r\) was \(m\) -_-

Answer 55

What if I tell you \(\large \mathbb{Z}[m,r]\)?