Three party-goers 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 party-goers in full and complete sentence.Make sure you reference any theorems that support your justifications.
Interesting party, what with all these nerds >:) ...well, some nerds they are XD What are your thoughts so far? Any ideas?
Haha well actually I don't get how to do this I'm so lost
Okay, consider the Professor's theory first. If (x+2) is a factor of a function, does it automatically mean that 2 is a zero of the function?
Yes?
It does, does it? Consider h(x) = (x+2)(x+3) And let x = 2. what is h(x)?
X+5x+6
Yes, but x = 2, and what happens?
You put 2 in every x so it would look like 2+10+6
Is that zero? :P
2 + 10 + 6 is that zero?
No
No. So 2 wasn't a zero of h(x)... was it? :P
2+10+6
18. Not zero. So 2 was not a zero of h(x), since h(2) is not zero. Understood?
Yes !
So, the professor is wrong...
Ok so what about ms guerra
Ms. Guerra gave the precise definition of a zero of a function. Can't argue with her :P
What about mr Romano?
Ever synthetic division?
We didn't use synthetic division
That's a problem. Synthetic division with r gives you the result when a polynomial is divided by (x-r)
Ok well I know how to somewhat do that
So do I divide (x+2) by something?
No, all I mean is that synthetic division with 2 means it was divided by (x-2) and there was no remainder, right?
Yes
So, that means g(x) = (x-2)f(x) for some function f(x)
Ok that it?
Well, try having x=2 now.
If I did this I got 1+3
How?
Ok we'll if I understood this right I plugged in my f(x) equation you showed me and I divided it by (x+2) did I do it wrong? Sorry I'm such a pain...
You're not a pain, lol But in g(x) = (x-2)f(x) Just replace all the x's with 2. Show me what happens...step-by-step if you must ^_^
Ok plug in 2 in all x's I got 0 f(2)
0 times f(2) 0 times anything is 0 So you got g(2) equals zero after all.
Ooo ok now I got what you where saying
So, by the third nerd's reasoning, yes, it is a valid argument. Meaning it's only (ironically) the professor who's off. Got it? good :P
Than you so much for helping me I have you a medal ;)
No problem. Just polish it up and you'll be in shape ^_^
gave* sorry and um thank you again.
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