#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.

8x^3+1

and...

x^7+y^7

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

#2

i hate algebra

#3

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.

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

#4

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...

#5

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.

#6

(x^3+y^4)(x^4+y^3) would that be it?

#7

= 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

#8

= 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

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?

#10

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.

#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

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

PENGUINEDIT: To Muphin, yeah, not trying to be a jerk or anything, just trying to help.

```
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.

PENGUINEDIT: To Muphin, yeah, not trying to be a jerk or anything, just trying to help.

#13

= 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.

#14

I don't think you can ever factor an ecuation or expression with only 2 terms and only one incognite...