2006-07-27 15:26:59 · 7 answers · asked by ToppDawg 2 in Science & Mathematics ➔ Mathematics
(3p)^7+7((3p)^6*(-2)) + 21((3p)^5*(-2)^2) + 35((3p)^4*(-2)^3) + 35((3p)^3*(-2)^4) + 21((3p)^2*(-2)^5) + 7((3p)*(-2)^6) + (-2)^7
2006-07-27 16:52:28 · answer #1 · answered by sanjeewa 4 · 4⤊ 1⤋
It would have helped if you explained your circumstances better. Just so we are clear, you want to expand the bynomial (3p-2) raised to the power of 7. You could do it long-hand by simple multiplication.
However, a more fun way of doing it is using Pascal's triangle to do the bynomial expansion. I have included a link in the sources for a site that explains this very well. Good luck.
2006-07-27 15:37:59 · answer #2 · answered by Nelson G 2 · 0⤊ 0⤋
3p^7 - 2^7
2006-07-27 15:29:12 · answer #3 · answered by losersrus 3 · 0⤊ 0⤋
2187 p^7 - 10206 p^6 + 20412p^5 - 22680 p^4 +15120 p^3 - 6068 p^2 + 1344 p -128
2006-07-27 15:31:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Use the general binomial-expansion theorem: (a + b)^n = \sum_{k=0}^n bc(n-choose-k) a^k b^(n-k), where bc(n-choose-k) is a binomial coefficient, which is equal to n!/(k! (n-k)!), where n! is equal to the product: n(n-1)(n-2)(n-3) ... (3)(2)(1) and so on. For more explanation click on the link I put under "Sources". It points to a wikipedia page on the binomial-expansion theorem.
2006-07-27 15:36:08 · answer #5 · answered by pollux 4 · 0⤊ 0⤋
2187p^7-10206p^6+20412p^5-22680p^4+15120p^3-6048p^2+1344p-128
2006-07-27 15:40:08 · answer #6 · answered by trackdude04 1 · 0⤊ 0⤋
clueless!!!!!
2006-07-27 15:29:17 · answer #7 · answered by va8326 5 · 0⤊ 0⤋