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Mathematics 20 Online
OpenStudy (anonymous):

2(x+1)^2 =2(x+1)(x+1)???

OpenStudy (anonymous):

you realize that we have (x+1)^2 on both sides right?

OpenStudy (anonymous):

thats not my answer, i just want to know if 2(x+1)^2 can be written as2(x+1)(x+1)

OpenStudy (anonymous):

then perhaps you should specify that in your question. And yes it can be written that way

OpenStudy (anonymous):

could you tell me the answer?

OpenStudy (anonymous):

what answer? the two sides are equal

OpenStudy (anonymous):

so is the who term raised to the 2nd power, or just the (x-1)??

OpenStudy (anonymous):

no only the (x+1) term is

OpenStudy (anonymous):

so how can they both be equal????

OpenStudy (anonymous):

(x+1)^2=x^2+2x+1 (x+1)(x+1)=x^2+2x+1

OpenStudy (anonymous):

it is 2(x+1)^2

OpenStudy (anonymous):

what i wrote above tells you that both sides of the equation are equal you can write it either way

OpenStudy (anonymous):

I am confused, what I did is; 2(x+1)^2 =2(x+1)(x+1) =2x+2(x+1)

OpenStudy (anonymous):

let me write it this way: 2(x+1)^2=2(x^2+2x+1). 2(x+1)(x+1)=2(x^2+2x+1) We have to take care of the parentheses first.

OpenStudy (anonymous):

did you distribute the 2 through both parentheses?

OpenStudy (anonymous):

you can if you want to

OpenStudy (anonymous):

but is that what you are suppose to do?

OpenStudy (anonymous):

what did you do???

OpenStudy (anonymous):

depends on what the directions say, but if i were you i wuld distribute the 2 into the parentheses

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

I got 2x^2+4x+2????

OpenStudy (anonymous):

nice job, that is the correct way to distribute the 2

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