(3+2x)^5
= 3^5 + 5(3^4)(2x) + 10(3^3)(2x)^2 + 10(3^2)(2x)^3 + 5(3)(2x)^4 + (2x)^5
=243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5
by simple observation that a negative base raised to an odd exponent is negative ....
(3-2x)^5
= 243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
(3+2x)^5 +(3-2x)^5
= 243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5 + 243 - 810x + 1080x^2 - 720x^3 + 240x^4 - 32x^5
= 486 + 2160x^2 + 480x^4
confirmed by Wolfram
http://www.wolframalpha.com/input/?i=expand+%283%2B2x%29%5E5+%2B%283-2x%29%5E5
Find
1. the binomial expansion of (3+2x)^5 simplifying terms. Hence find the binomial expansion of (3+2x)^5 +(3-2x)^5.
1 answer
1 answer