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Mathematics 15 Online
kekeman:

I need help with math

kekeman:

1 attachment
axie:

@vocaloid

Gucchi:

is this for an exam?

Joe348:

I believe for the first part it would be -5 and something

kekeman:

No its not for an exam

Gucchi:

whats this?

1 attachment
kekeman:

Its for my segment exam notes

kekeman:

@joe348 wrote:
I believe for the first part it would be -5 and something
Hmmmm ok thanks

lifewmylilweeb:

@joe348 wrote:
I believe for the first part it would be -5 and something
yeah there is (-5,0), (-1,0), (4,0), and (7,0)

kekeman:

Thanks

lifewmylilweeb:

yw

lea0728:

@lazorwolf64

kekeman:

@lifewmylilweeb wrote:
@joe348 wrote:
I believe for the first part it would be -5 and something
yeah there is (-5,0), (-1,0), (4,0), and (7,0)
Wait there is no -2?

SmokeyBrown:

@kekeman wrote:
Wait there is no -2?
When x = -2, y is around 4, not 0, so -2 would not be a zero of the curve

kekeman:

Ohhhh ok thanks

kekeman:

I just need to figure out what to come up with on part B.)

SmokeyBrown:

For part 2 of the question, you don't need to give the actual equation of the curve, just a possible equation based on the zeroes you found in part 1. (By the way, I think -5 and 4 are the ones with "possible multiplicities", since the shapes when they intersect are a parabola and a cubed function, respectively. I could be off base with this conclusion) So, with the zeroes -5, -1, 4, 7, assuming -5 has multiplicity of 2 and 4 has multiplicity of 3, one possible equation could be: (x+5) * (x+5) * (x+1) * (x-4) * (x-4) * (x-4) * (x-7) This would also result in a seventh-degree polynomial, as the problem describes since 7 terms are multiplied together. I think this is what the question is getting at, but to be honest, I'm guessing a bit on some parts

kekeman:

@smokeybrown wrote:
For part 2 of the question, you don't need to give the actual equation of the curve, just a possible equation based on the zeroes you found in part 1. (By the way, I think -5 and 4 are the ones with "possible multiplicities", since the shapes when they intersect are a parabola and a cubed function, respectively. I could be off base with this conclusion) So, with the zeroes -5, -1, 4, 7, assuming -5 has multiplicity of 2 and 4 has multiplicity of 3, one possible equation could be: (x+5) * (x+5) * (x+1) * (x-4) * (x-4) * (x-4) * (x-7) This would also result in a seventh-degree polynomial, as the problem describes since 7 terms are multiplied together. I think this is what the question is getting at, but to be honest, I'm guessing a bit on some parts
Polynomial's zeroes: -5, -1, 4, 7 Possible multiplicities: x = 7 (multiplicity of 1), x = 4 (multiplicity of 1), x = -1 (multiplicity of 1), x = -5 (multiplicity of 2). Is this right?

SmokeyBrown:

I think 4 has a multiplicity of 3, based on the shape of the curve near that point, but I could be mistaken about that

kekeman:

Omg yes thats what i was thinking

kekeman:

So 4 has a multiplicity of

kekeman:

3

kekeman:

Which one would be a better one for part B.) (x+5)^2(x+1)(x+2)(x-4)(x-7) or -(x-4)^3*(x+5)^2*(x+1)*(x-7)

SmokeyBrown:

Sorry for the late response I think I would go with the second option, -(x-4)^3*(x+5)^2*(x+1)*(x-7) The formatting is a bit neater and it shows the proper multiplicity of 4 and -5

kekeman:

Ok even putting "*" in between it?

SmokeyBrown:

Yeah, you can use the asterisk * to signify multiplication so that's fine to include

kekeman:

Ok thanks :)

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