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Mathematics 15 Online
OpenStudy (anonymous):

Help. Help. Help

OpenStudy (anonymous):

:O

OpenStudy (anonymous):

Can you be obsessed with my question? lmao

OpenStudy (anonymous):

Mhm

OpenStudy (anonymous):

Find the zeros of each function. State the multiplicity of multiple zeros. 53) (x+2)^2(x-5)^4

OpenStudy (anonymous):

what do you need to plug into x to get zero?

OpenStudy (anonymous):

it's given away in the parenthesis

OpenStudy (anonymous):

-2?

OpenStudy (anonymous):

that's one of them :)

OpenStudy (anonymous):

5?

OpenStudy (anonymous):

so what does multiplicity mean?

OpenStudy (anonymous):

sorry, I had to look it up :) it's how many zero's there are

OpenStudy (anonymous):

Ok so (-2+2)^2(5-5)^4

OpenStudy (anonymous):

no, sorry :)

OpenStudy (anonymous):

when it's (x+2)^2 it's really (x+2)*(x+2) and (x-5)^4 is really (x-5)*(x-5)*(x-5)*(x-5)

OpenStudy (anonymous):

so there are 6 places that there can be a zero :)

OpenStudy (anonymous):

does that help?

OpenStudy (anonymous):

omg yes

OpenStudy (anonymous):

Good :) take care

OpenStudy (anonymous):

waaaaait

OpenStudy (anonymous):

Let me verify

OpenStudy (anonymous):

ok go

OpenStudy (anonymous):

51) f(t)=4t+3t^3+2t-7 4t + 3t(3t)(3t) +2t-7 5 possible zeros?

OpenStudy (anonymous):

I'm sorry friend :? I gotta run.. Just do the same thing you did in the last one!! look for what will make it zero!! Good luck :D

OpenStudy (anonymous):

Okay ty take care

OpenStudy (anonymous):

you too

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