#1
So I'm in algebra 2 and ****... I havn't done this stuff for 2 years so I feel really stupid but I forgot how to do these simple factoring problems...
8x^3+1
and...
x^7+y^7

^=to the power of BTW so 5^2 is 5 to the second power.
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#2
i hate algebra
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#4
Quote by ibryant0915
So does the problem say to 'factor' it or 'solve for x' or what?

As it is right now you can't factor those any further... unless I'm just being idiotic.

Hmm. It says to factor it...
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#5
Quote by soul_power
So I'm in algebra 2 and ****... I havn't done this stuff for 2 years so I feel really stupid but I forgot how to do these simple factoring problems...
8x^3+1
and...
x^7+y^7

^=to the power of BTW so 5^2 is 5 to the second power.


= 8x^3+1
= (2x+1)^3

= x^7+y^7
= (x+y)^7 <=== This doesn't seem right to me, please correct.
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#7
Quote by Muphin
= 8x^3+1
= (2x+1)^3

= x^7+y^7
= (x+y)^7 <=== This doesn't seem right to me, please correct.


Actually i think your sign in the () of #1 needs to be a negative. I could be wrong though,i havent worked it out, just looked at it
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#8
Quote by Muphin
= 8x^3+1
= (2x+1)^3

= x^7+y^7
= (x+y)^7 <=== This doesn't seem right to me, please correct.

I think you're wrong.
(2x+1)^3 = (2x+1)(2x+1)(2x+1) = (4x^2 + 4x + 1)(2x+1) = 8x^3 + 12x^2 +4x + 1

(x+y)^7 = (x+y)(x+y)(x+y)(x+y)(x+y)(x+y)(x+y) and you get:
x^7 + 7(x^6)y + 21(x^5)(y^2) + 35(x^4)(y^3) + 35(x^3)(y^4) + 21(x^2)(y^5) + 7(x)(y^6) + y^7
(using Pascal's Triangle)
#9
Quote by gibsonpenguin
I think you're wrong.
(2x+1)^3 = (2x+1)(2x+1)(2x+1) = (4x^2 + 4x + 1)(2x+1) = 8x^3 + 12x^2 +4x + 1

(x+y)^7 = (x+y)(x+y)(x+y)(x+y)(x+y)(x+y)(x+y) and you get: x^7 + 7(x^6)y + 21(x^5)(y^2) + 35(x^4)(y^3) + 35(x^3)(y^4) + 21(x^2)(y^5) + 7(x)(y^6) + y^7 (using the triangle thing I forget the name of.)

Pythagorean theorem?
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#10
Quote by gibsonpenguin
I think you're wrong.
(2x+1)^3 = (2x+1)(2x+1)(2x+1) = (4x^2 + 4x + 1)(2x+1) = 8x^3 + 12x^2 +4x + 1


Good call.
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#11
I'm pretty sure that neither of those are factorable. Is "not factorable" a choice?
#12
I googled it, it's Pascal's Triangle, you use it to expand factor-y things


          1
         1 1
        1 2 1
       1 3 3 1
      1 4 6 4 1
    1 5 10 10 5 1
   1 6 15 20 15 6 1
  1 7 21 35 35 21 7 1


Not the best formatting ever, but you see how it works.

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#13
Quote by Muphin
= 8x^3+1
= (2x+1)^3

= x^7+y^7
= (x+y)^7 <=== This doesn't seem right to me, please correct.


It's not, it would turn out to something like

x^7+yx^6+y^2x^5 etc.

I don't really know. It looks factored to me already.

^yeah that's totally wrong but whatever.