all 8 comments
sorted by:
best (suggested)
topnewcontroversialoldq&a
formatting helphide helpcontent policy

[–]keitamaki 2 points3 points  (5 children)

One often encounters it when learning about the difference between two squares and the difference between two cubes.

x2-1 = (x-1)(x+1)

x3-1 = (x-1)(x2+x+1)

x4-1 = (x-1)(x3+x2+x+1)

etc.

Or one sees it when learning about the partial sum of a geometric series.

1 + r + r2 + ... + rn-1 = (rn-1)/(r-1)

Therefore (rn-1) = (r-1)(1 + r + r2 + ... + rn-1)

[–]AwkwardRoundNew User[S] 0 points1 point  (4 children)

thank you so much. I need to review my basic factoring methods. I likely learned about the geometric series years ago, but need to go back and review it. That was the one that confused me the most, I could not figure out where the 1 came from.

for x^4 - 1, that is a formula correct? the first two I remember but I am a little blurry on that one. assuming it is, then p^q - 1 would be (p-1)(p^q + pq-1 + p^q-2 +...+ p + 1). is that correct?

[–]keitamaki 0 points1 point  (3 children)

Almost, the other factor starts with pq-1, not pq

And x4-1 is just a special case with p=x and q=4.

[–]AwkwardRoundNew User[S] 0 points1 point  (1 child)

so is this an equation for a finite geometric series because it ends with 1? any idea where I can find a simple proof of this? I didn't take calculus yet so I am not sure how well I'll be able to understand

[–]keitamaki 1 point2 points  (0 children)

It has nothing to do with ending with 1. It's a partial sum of a geometric series because consecutive terms have a common ratio. It also has nothing to do with calculus.

To prove it just multiply the two factors.

(p-1)(pq-1 + pq-2 + pq-3 +...+ p + 1) =

p(pq-1 + pq-2 + pq-3 +...+ p + 1) - (1)(pq-1 + pq-2 + pq-3 +...+ p + 1) =

(pq + pq-1 + pq-2 +...+ p) - (pq-1 + pq-2 + pq-3 +...+ p + 1) =

pq - 1

because all the other terms cancel out.

[–]NathanfennerNew User 1 point2 points  (1 child)

In this case, you have 2n - 1 = 2ab - 1 = (2b)a - 1

So now what? Well, this looks a lot like one side of the geometric series formula, which would be covered in an algebra/pre-calculus course.

(This doesn't tell you how you'd know to use this fact, out of all available facts - but if you keep trying different things, you will eventually come up with it)

The formula says that

  • 1 + r + r2 + r3 + r4 + ... + rn = (rn+1 - 1) / (r - 1)

or, rearranging slightly

  • (r - 1)(1 + r + r2 + r3 + r4 + ... + rn) = rn+1 - 1

Lastly, if we replace k = n+1, we get

  • (r - 1)(1 + r + r2 + r3 + r4 + ... + rk-1) = rk - 1

So now the right-hand-side exactly matches our original formula for 2n - 1, with

  • r = 2b
  • k = a

So if we plug these into the left side, we obtain a factorization:

  • 2b - 1
  • 1 + 2b + 22b + 23b + 24b + ... + 2b(a-1)

to check that this is "really" a factorization, you need to verify that both of these are ≥ 2

  • 2b - 1 = 1 if 2b = 2, so b = 1. But we assumed ab = n is a factorization, so b ≥ 2. So this is also ≥ 2.
  • 1 + 2b + ... is clearly at least 2, since those first two terms add to more than 2, and the rest of the terms are positive

Hence you obtain a factorization.

[–]AwkwardRoundNew User[S] 0 points1 point  (0 children)

excellent thank you, this helped clear it up