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Mathematics 16 Online
OpenStudy (anonymous):

Three partygoers are in the corner of the ballroom having an intense argument. You walk over to settle the debate. They are discussing a function g(x). You take out your notepad and jot down their statements. Professor McCoy: She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0. Ms. Guerra: She says that 2 is a zero of g(x) because g(2) = 0. Mr. Romano: He says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0. Correct the reasoning of any inaccurate reasoning by the partygoers in full and complete sentences.

OpenStudy (anonymous):

Please help I dont get this. I will reward a medal and fan you!

OpenStudy (anonymous):

@leon549

OpenStudy (anonymous):

@dahlindubs11

OpenStudy (anonymous):

@israa88

OpenStudy (anonymous):

first one is right

OpenStudy (anonymous):

Can you tell me why?

OpenStudy (anonymous):

No, I think the last two are correct. Do you know why?

OpenStudy (anonymous):

because he made that choice after solving the hole problem

OpenStudy (anonymous):

Prof. McCoy should have said (x-2) not (x+2).

OpenStudy (anonymous):

second one is wrong because he just put the value of x in the eq without solving the eq

OpenStudy (anonymous):

The factor theorem says that: If f(a)=0, then (x-a) is a factor The remainder theorem says that: If (x-a) is a factor of f(x), then f(x)x−a=0

OpenStudy (anonymous):

yeah i din't see that

OpenStudy (anonymous):

So @LexiLuvv2431 : Did you understand anthing I said?

OpenStudy (anonymous):

@LexiLuvv2431: So say that 2 is indeed a zero of f(x), then a factor must be (x-2) according to the first which then supports Ms. Guerra and also if (x-2) really is a factor as Ms. Guerra says then we can understand that f(x)x−2=0 which supports Mr. Romano conjecture.

OpenStudy (anonymous):

Does this help?

OpenStudy (anonymous):

Yess I kinda get it. So is Ms. Guerra right?

OpenStudy (anonymous):

Yes. And so is Mr.Romano.

OpenStudy (anonymous):

Ohh okayy I see. Thank you!

OpenStudy (anonymous):

Anytime. :)

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