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May 6th, 2008, 07:49 PM  #1 
Newbie Joined: May 2008 Posts: 2 Thanks: 0  I need help with raising a polynomial to some power
these are my problems: (D+3)^14 (3X+4Y)^7 
May 6th, 2008, 08:56 PM  #2  
Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0 
Have you ever heard of Pascal's Triangle? It has several unusual and interesting properties. I think that if I were asked to raise a polynomial to the nth power, I would use Pascal's Triangle. To form, start with a row of length one and with one value, '1' Each subsequent row has an additional value, and every value is the sum of the two values above it (except the second row, which only has one value above it). For example, Quote:
First row = 1 Second row: above each value is only one entry, so again each value is 1 Third row: Above the middle value are the two integers 1 and 1, so we have '2' Fourth row: the second value is below 1 and 2, so we have '3' and so on Now for why it's useful. The values for Pascal's Triangle reflect the coefficients of a binomial expansion. Say we have (A + B) raised to some power. (A + B)^0 = 1, the first row of Pascal. (A + B)^1 = A + B; the coefficients are the second row of Pascal. (A + B)Â² = AÂ² + 2AB + BÂ²; 1 2 1, the coefficients are the third row. (A + B)Â³ = AÂ³ + 3AÂ²B + 3ABÂ² + BÂ³ Say we wanted (A + B)^5. Well, the sixth row of Pascal's Triangle is 1 5 10 10 5 1 So without doing anything other than that bit of addition, I know that it must equal A^5 + 5A^4*B + 10AÂ³BÂ² + 10AÂ²BÂ³ + 5AB^4 + B^5 And as an additional note, let's say we have (2X + Y)Â² Well, Pascal says our coefficients are 1 2 1. Letting 2x be A, and since we know (A + B)Â² = AÂ² + 2AB + BÂ², we know that it must equal (2X)Â² + 2(2X)(Y) + (Y)Â² Or you could just multiply it out. Whichever you prefer.  
May 6th, 2008, 09:09 PM  #3 
Newbie Joined: May 2008 Posts: 2 Thanks: 0  I need answers cherry cakes
I need answers cherry cakes

May 6th, 2008, 09:11 PM  #4  
Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0  Re: I need answers cherry cakes Quote:
 
May 7th, 2008, 08:25 AM  #5 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
You could download Maxima or another symbolic algebra system if you wanted just the answer. Jason was just trying to explain how to answer the problem  much more important, I think, than the actual answer.


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