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Mathematics 17 Online
OpenStudy (anonymous):

IMPORTANT! Find all the zeros of each function. State the multiplicity of any multiple zeros. y=(x-1)^2(2x-3)^3 = 1 has a multiplicity of 2 & 2/3 has a multiplicity of 3? This one I have no idea: y=(x+2)(x^2+3x-40)

OpenStudy (anonymous):

y = (x+2)(x²+3x-40) (x+2)(x²+3x-40)=0 x+2=0 or x²+3x-40=0 then we have x = -3 +- srqrt169 / 2 x = -3 +- 13 / 2 x' = -16/2 and x'' = 10/2 finally... x = - 2, x = -8 and x = 5. is that what you mean?

OpenStudy (anonymous):

so i got the first question i asked right? &

OpenStudy (anonymous):

How did you get then we have x = -3 +- srqrt169 / 2?

OpenStudy (anonymous):

Sorry, I don't know what you mean.

OpenStudy (anonymous):

mmm lemme see

OpenStudy (anonymous):

its Bhaskara

OpenStudy (anonymous):

from x² + 3x - 40

OpenStudy (anonymous):

Can you explain to me how you did that step please

OpenStudy (anonymous):

quadratic formula

OpenStudy (anonymous):

you do that with x² + 3x - 40 (ax² +bx + c)

OpenStudy (anonymous):

just replace the letters

OpenStudy (mertsj):

\[x^2+3x-40=(x+8)(x-5)\]

OpenStudy (dumbcow):

you can also factor this: x^2 +3x -40 = (x+8)(x-5)

OpenStudy (mertsj):

No need to us the quadratic formula here.

OpenStudy (anonymous):

oh that's right, easier indeed.

OpenStudy (anonymous):

Can you explain to me how you factored that

OpenStudy (dumbcow):

@cubanito for your answer posted, the 2nd zero should be 3/2 not 2/3 2x-3 = 0 2x = 3 x = 3/2

OpenStudy (anonymous):

Yeah, I meant to put 3/2. But those two answers are right?

OpenStudy (anonymous):

cubanito, you can factor that like this: x² + 3x - 40 x 8 x -5

OpenStudy (anonymous):

So, its x = - 2, x = -8 and x = 5?

OpenStudy (anonymous):

Yep, it is. But make sure you got the whole process, not only the answers.

OpenStudy (anonymous):

What about y= x^3+x^2-19x+5

OpenStudy (anonymous):

You want to find the value of x?

OpenStudy (anonymous):

no, find all the zeros & state the multiplicity

OpenStudy (dumbcow):

try zeros -1,-2,-3,-4,-5 and see if they factor out evenly i suggest using synthetic division

OpenStudy (anonymous):

Yea, that's a good way, dumbcow... wait, dumbcow? wth is this name? LOL

OpenStudy (dumbcow):

i like the irony of it :) also if you have a graphing calculator, graph the function and see where it crosses x_axis

OpenStudy (anonymous):

where does it cross the x axis

OpenStudy (anonymous):

when y =0

OpenStudy (anonymous):

I'm confused now :(

OpenStudy (anonymous):

Melinda....dumbcow?????!!!???

OpenStudy (anonymous):

moved here http://openstudy.com/updates/4f43a67fe4b065f388dcafc9

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