Because the Binomial Theorem involves the addition of two terms raised to a power, we rewrite (2 x-y)^5 as [2 x+(-y)]^5. We use the Binomial Theorem

(a+b)^n=\left(\begin{array}{l}n \\0\end{array}\right) a^n+\left(\begin{array}{l}n \\1\end{array}\right) a^{n-1} b+\left(\begin{array}{l}n \\2\end{array}\right) a^{n-2} b^2+\left(\begin{array}{l}n \\3\end{array}\right) a^{n-3} b^3+\cdots+\left(\begin{array}{l}n \\n\end{array}\right) b^n

to expand [2 x+(-y)]^5. In [2 x+(-y)]^5, a = 2x, b = -y, and n = 5. In the expansion, powers of 2x are in descending order, starting with (2 x)^5. Powers of -y are in ascending order, starting with (-y)^0. [Because (-y)^0 = 1, a – y is not shown in the first term.] The sum of the exponents on 2x and -y in each term is equal to 5 , the exponent in the expression (2 x-y)^5.

(2 x-y)^5=[2 x+(-y)]^5

=1\left(32 x^5\right)+5\left(16 x^4\right)(-y)+10\left(8 x^3\right) y^2+10\left(4 x^2\right)\left(-y^3\right)+5(2 x) y^4+1\left(-y^5\right)

Multiplying factors in each of the six terms gives us the desired expansion:

(2 x-y)^5=32 x^5-80 x^4 y+80 x^3 y^2-40 x^2 y^3+10 x y^4-y^5.