Precalculus, Chapter 9, 9.5, Section 9.5, Problem 24

You need to use the binomial formula, such that:
(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k
You need to replace y for x, 2 for y and 5 for n, such that:
(y-2)^5 = 5C0 (y)^5+5C1 (y)^4*(-2)^1+5C2 (y)^3*(-2)^2+5C3 y^2*(-2)^3 + 5C4 y*(-2)^4 + 5C5 (-2)^5
By definition, nC0 = nCn = 1, hence 5C0 = 5C5 = 1.
By definition nC1 = nC(n-1) = n , hence 5C1 = 5C4 = 5.
By definition nC2 = (n(n-1))/2 , hence
(y-2)^5 = y^5 - 10y^4+40y^3- 80y^2 + 80y - 32
Hence, expanding the complex number using binomial theorem yields the simplified result (y-2)^5 = y^5 - 10y^4+40y^3- 80y^2 + 80y - 32