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Mathematics 14 Online
OpenStudy (help_people):

Which polynomial is a difference of two squares? x2 − 16 x2 + 16 x2 − 8 x2 + 8

OpenStudy (help_people):

@amistre64 @Mehek14

OpenStudy (dinamix):

its first one

OpenStudy (jhannybean):

According to what we know about polynominals, let's say we take \(x=a\) and \(4 =b\) Then we can take our form, \(a^2-b^2\) to write out our function as \((x)^2-(4)^2\) which simplifies to.....?

OpenStudy (dinamix):

\[x^2-16 = (x-4)(x+4)\]@help_people and \[x^2-8 = (x-\sqrt{8})(x+\sqrt{8})\]

OpenStudy (help_people):

ok @Jhannybean and @dinamix I know the answer but could you explain more for osme reason i do not get it

OpenStudy (zzr0ck3r):

@help_people You re looking for the form \(a^2-b^2\) and the things is, when it is a variable, we can have it be \(x^2\) so its the other part we are concerned with. So since we need a \(-\) sign, we can rule out options 2 and 4. Ok so what is the problem with \(x^2-8\) Well in order for it to be a perfect square, we need the thing after the \(-\) sign to be written as \(b^2\) where \(b\) is a INTEGER. \(\sqrt{8}\) is not an integer, but \(\sqrt{16}=4\) is.

OpenStudy (zzr0ck3r):

so we have \(x^2-16=x^2-4^2\)

OpenStudy (help_people):

ok

OpenStudy (dinamix):

@help_people u study number complexe , @zzr0ck3r

OpenStudy (dinamix):

not you i am sopke with him , i ask if study it

OpenStudy (zzr0ck3r):

I am not sure what you are asking. I am good in certain area dealing with complex numbers, but not the traditional. There are much better people here for that.

OpenStudy (jhannybean):

I really like your explanation here, @zzr0ck3r

OpenStudy (zzr0ck3r):

hi @Jhannybean

OpenStudy (zzr0ck3r):

ty :)

OpenStudy (dinamix):

@zzr0ck3r i hope know how to explain like u

OpenStudy (zzr0ck3r):

often my problem is over explaining. I have to learn to say only the things that are important. Some stuff I said up there is not 100% correct, but if I say it 100% correct often I would confuse everyone.

OpenStudy (help_people):

ok thank you for everthing :/ @zzr0ck3r XOXO to all

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