`(y - 2)^5` Use the Binomial Theorem to expand and simplify the expression.

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`