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Mathematics 8 Online
OpenStudy (anonymous):

Pleeeeaaase help!! I will become a fan and give a medal to best answer! What is the product? x^2-3x+2/8x+16 * 4/x^2-1 State any restrictions on the variable. Please show all steps!

OpenStudy (anonymous):

restrictions are the zeros of the denominator first you need to factor each expression and from that you can easily cancel out the terms that are alike.. then simplify

OpenStudy (anonymous):

restrictions are the zeros of the denominator first you need to factor each expression and from that you can easily cancel out the terms that are alike.. then simplify

OpenStudy (anonymous):

wow you're thinking what i'm thinking :) welcome to OS @justicenina (^_^)

OpenStudy (anonymous):

(x-1)(x-2)/8(x-2) * 4/(x-1)(x+1)

OpenStudy (anonymous):

Is that right?

OpenStudy (anonymous):

almost right :) check the denominator of the first rational function it should be 8(x + 2) not (x-2)

OpenStudy (anonymous):

thank you Data_LG2

OpenStudy (anonymous):

So then would it be (x-2)/8(x+2) * 4/(x+1) ?

OpenStudy (anonymous):

@Data_LG2

OpenStudy (anonymous):

if I help you now will I get the medal and fan?

OpenStudy (anonymous):

Lol yes if you really know how to do this

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

K give me 2 min let me get my pen

OpenStudy (anonymous):

answer (x-2)/2(x+2)(x+1) x can not be 1

OpenStudy (anonymous):

Can you show me the steps you did?

OpenStudy (anonymous):

Let me know if you follow this (x-1)(x-2)/8(x+2) * 4/(x-1)(x+1) Now you cancel (x-1) term and 4 and 8

OpenStudy (anonymous):

So after cancelling it would be (x-2)/(x+2) * (x+1) ?

OpenStudy (anonymous):

NO 4/8 = 1/2

OpenStudy (anonymous):

(x-2)/2(x+2) * 1/(x+1) ?

OpenStudy (anonymous):

YES

OpenStudy (anonymous):

answer (x-2)/2(x+2)(x+1) x can not be 1 Where you wrote this is it supposed to be (x-2)/2(x+2) * (x+1) ?

OpenStudy (anonymous):

@thisSucks

OpenStudy (anonymous):

Yes it is (x-2)/2(x+2) * (x+1)

OpenStudy (anonymous):

So that's all the work for this problem?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

np :)

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