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May 6th, 2008, 07:49 PM   #1
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I need help with raising a polynomial to some power

these are my problems:

(D+3)^14

(3X+4Y)^7
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May 6th, 2008, 08:56 PM   #2
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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:
Originally Posted by Pascal's Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
(pretend it made an equilateral triangle, not a right triangle)

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)² = A² + 2AB + B²; 1 2 1, the coefficients are the third row.
(A + B)³ = A³ + 3A²B + 3AB² + B³

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 + 10A³B² + 10A²B³ + 5AB^4 + B^5

And as an additional note, let's say we have (2X + Y)²
Well, Pascal says our coefficients are 1 2 1. Letting 2x be A, and since we know (A + B)² = A² + 2AB + B², we know that it must equal (2X)² + 2(2X)(Y) + (Y)²

Or you could just multiply it out. Whichever you prefer.
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May 6th, 2008, 09:09 PM   #3
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I need answers cherry cakes

I need answers cherry cakes
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May 6th, 2008, 09:11 PM   #4
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Re: I need answers cherry cakes

Quote:
Originally Posted by amboo
I need answers cherry cakes
Perhaps, little baker, you could place the cherries on top yourself. But if you need help, I suspect that we'd be very capable.
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May 7th, 2008, 08:25 AM   #5
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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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