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Postby JustcallmeDrago » Mon Oct 29, 2007 2:48 am

I see Ike is on here.

Hey Ike,

how would you factor x^4-(a^2)(x^2)+16a^4 ?

It looks like a simple binomial squared but the " -a2x2 " has me stumped. how can you possibly get this?
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Postby Ikerous » Mon Oct 29, 2007 3:01 am

If you take u=x^2 and w=a^2 you get u^2 - wu + 16w^2

That doesn't seem factorable :/

I'll keep looking at it though
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Postby JustcallmeDrago » Mon Oct 29, 2007 3:06 am

Ikerous wrote:If you take u=x^2 and w=a^2 you get u^2 - wu + 16w^2

That doesn't seem factorable :/

I'll keep looking at it though


There are others, but that is the most irritating one...
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Postby Ikerous » Mon Oct 29, 2007 3:09 am

You can post them. I might fail less at those than this one.
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Postby JustcallmeDrago » Mon Oct 29, 2007 3:17 am

Ikerous wrote:You can post them. I might fail less at those than this one.


=D yay!

x^8 + x^3

I got x^3(x^5 +1), but that just feels wrong.


Same with x4y2 + 2x3y2 + x2y2 - x2y4

I factored out x2y2 and got x2y2(x2 + 2x + 1 - y2), which is ugly :{

And I don't even know where to begin with x^(2a+1) + x^(a+1) - 6x

Answer/don't answer any you want (or don't want) to.
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Postby Ikerous » Mon Oct 29, 2007 3:22 am

JustcallmeDrago wrote:
Ikerous wrote:You can post them. I might fail less at those than this one.


=D yay!

x^8 + x^3

I got x^3(x^5 +1), but that just feels wrong.


Same with x4y2 + 2x3y2 + x2y2 - x2y4

I factored out x2y2 and got x2y2(x2 + 2x + 1 - y2), which is ugly :{

And I don't even know where to begin with x^(2a+1) + x^(a+1) - 6x

Answer/don't answer any you want (or don't want) to.

I'd say that's as factored as the first one gets. The second one might be able to go a little farther with x^2 y^2 [(x+1)^2 - y^2]
Which would go to x^2 y^2 (x+1-y)(x+1+y)

As for the last one you can factor out an x^(a+1) and get x^(a+1) [x^a + 1] - 6x
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Postby JustcallmeDrago » Mon Oct 29, 2007 3:34 am

for #3 I can't believe i didn't realize x^2a+1 is similar to a normal exponent.

as for #2, that really helped and i see a difference of squares in there. (well, i just saw you edited that in there... but yeah i did find that.

i feel i've done pretty good; i have done 75 of these so far.

how about -(x-y)^2 + c^2 ? nothing is obvious past simplifying... :/
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Postby Ikerous » Mon Oct 29, 2007 3:38 am

JustcallmeDrago wrote:how about -(x-y)^2 + c^2 ? nothing is obvious past simplifying... :/

You can turn it around and get c^2 - (x-y)^2 which makes (c-x+y)(c+x-y)
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Postby JustcallmeDrago » Mon Oct 29, 2007 3:40 am

Ikerous wrote:
JustcallmeDrago wrote:how about -(x-y)^2 + c^2 ? nothing is obvious past simplifying... :/

You can turn it around and get c^2 - (x-y)^2 which makes (c-x+y)(c+x-y)

Oh. :oops: :oops:
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Postby TSchultz » Mon Oct 29, 2007 4:10 am

In[1]:= Factor[x^4 - (a^2) (x^2) + 16 a^4]
Out[1]= (4 a^2 - 3 a x + x^2) (4 a^2 + 3 a x + x^2)
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:13 am

TSchultz wrote:In[1]:= Factor[x^4 - (a^2) (x^2) + 16 a^4]
Out[1]= (4 a^2 - 3 a x + x^2) (4 a^2 + 3 a x + x^2)


so how did you get that?
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Postby TSchultz » Mon Oct 29, 2007 4:13 am

Mathematica.
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:18 am

TSchultz wrote:Mathematica.


Is it free? or does it show you the steps?
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Postby TSchultz » Mon Oct 29, 2007 4:23 am

For me it was free. It's free for all BU students, and all students at my high school have BU ID's so we can get it. But otherwise it costs a fair bit (100+ depending on if you're a student or not I think).
And sorry :/ doesn't show the steps. But there's an add on package that does I think.
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Postby JustcallmeDrago » Mon Oct 29, 2007 4:24 am

TSchultz wrote:For me it was free. It's free for all BU students, and all students at my high school have BU ID's so we can get it. But otherwise it costs a fair bit (100+ depending on if you're a student or not I think).
And sorry :/ doesn't show the steps. But there's an add on package that does I think.


X|
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