The coefficient of x5in the expansion of (2x + 5)8is:

To find the coefficient of x^5 in the expansion of (2x + 5)^8, we can use the binomial theorem, which states that (a + b)^n = Σ (n choose k) * a^(n-k) * b^k, where the sum is from k = 0 to n.

In this case, a = 2x, b = 5, n = 8, and we want to find the coefficient of x^5, so we need to find the term where k = 5.

The binomial coefficient (n choose k) = n! / [k!(n-k)!], where "!" denotes factorial.

So, the coefficient of x^5 is given by:

(8 choose 5) * (2x)^(8-5) * 5^5

= (876 / 321) * 2^3 * x^3 * 5^5

= 56 * 8 * x^3 * 3125

= 358400x^3

But we want the coefficient of x^5, not x^3. To get this, we need to divide our result by x^3, which gives us:

358400

So, the coefficient of x^5 in the expansion of (2x + 5)^8 is 358400.