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April 16, 2014

Homework Help: Pre-Calc-please check

Posted by Bethany on Wednesday, January 4, 2012 at 12:54pm.

Please check this for me
expand (3z-2b)^5 using Binomial Theorem
(a+b)^n = n on top sigma in middle k=0 to right (n/k)a^(n-k)b^k

(3a-2b)^5 = 5 on top sigma in middle k=0 on bottom(5/k)(3a)^5-k(-2b)^k

(5/0)(3a)^5(-2b)^0+(5/1)(3a)^4(-2b)^1 +(5/2)(3a)^3(-2b)^2

+(5/3)(3a)^2(-2b)^3 + (5/4)(3a)^1(-2b)^4+(5/5)(2a)^0(-2b)^5

=243a^5-810a^4b+1080a^3b^2+-720a^2b^3+240ab^4-32b^5

5!/(5-5)!5!=5!/1-5!=5!/5!1

(5/0)=5!/(5-0)!0!=5!/5!0!=1

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