April 5th, 2017, 10:47 PM  #11 
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  I think I get it by adding the 2 to the 2?
Last edited by GIjoefan1976; April 5th, 2017 at 10:48 PM. Reason: Wrong number 
April 5th, 2017, 11:11 PM  #12 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out.  $(x  2)(x  2) = x*x  2x  2x + (2)(2) = x^2  4x + 4$ Does this help? Then, if we multiply by $(x  2)$ again, $(x^2  4x + 4)(x  2) = x^2 * x  2x^2  4x*x + (4x)*(2) + 4x + 4*(2) \\\\ = x^3  6x^2 + 12x  8$ Remember that you must multiply ever term in the first set of parenthesis by everything in the second. I think they call it FOIL. 
April 6th, 2017, 06:28 AM  #13 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
Looking back, what you were trying to do in your original post was to generalize a procedure that worked for cases where n = 2 to cases where n = 3. Generalization (and abstraction) are the engines of mathematical thought, but they can be tricky. You got caught up in a false generalization. Easy to do, which is why mathematical conjectures must be proved. Let's proceed in baby steps. $(a + b)^1 = a + b.$ $(a + b)^2 = (a + b)(a + b) = (a + b) * a + (a + b) * b =$ $a^2 + ba + ab + b^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.$ $(a + b)^3 = (a + b)(a + b)(a + b) = (a + b)(a + b)^2 =$ $(a + b)(a^2 + 2ab + b^2) = a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2) =$ $a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3 =$ $a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 =$ $a^3 + 3a^2b + 3ab^2 + b^3.$ It is hard to see a pattern there on which to generalize. It took mathematicians thousands of years to see it. But you can easily see that the middle terms are all products of powers of a and b with the powers of a decreasing by 1 and the powers of b increasing by 1 middle term by middle term. One breakthrough came by defining x^0 = 1. With that it became obvious that every term is a product of powers, starting with 1 times a^n times b^0, with each following term being a numerical coefficient times a to a reduced power times b to an increased power. The other breakthrough came with the definition of $0! = 1\ \text{and}\ n! = n * (n  1)!\ \text{for}\ n \in \mathbb N^+.$ It then became possible to prove that $(a + b)^n = \displaystyle \left ( \sum_{j=0}^n \dfrac{n!}{j! * (n  j)!} * a^{n  j} * b^j \right ).$ There is a pattern, but it is complex. Last edited by skipjack; April 6th, 2017 at 07:18 AM. 
April 6th, 2017, 10:49 AM  #14  
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Quote:
 
April 6th, 2017, 11:01 AM  #15  
Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10  Quote:
Yep that's exactly what happened! I thought it would work for 3 just as it did for for 2 but it wasn't so I was so lost. Thanks for your time, the information, and for the baby steps. I am very much still at the baby steps stage. No one here on this site really does it but some of the tutors at my school are way too quick for me and always in a rush, so I get scared to ask them to slow down so as always thanks everyone for helping me get through this, it really eats at my soul when I just stare at a problem for hours on end just getting angry at myself.  

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