What is the binomial expansion of (2x - 3) ^5?

A) (2x) ^ 5 - 15 (2x) ^ 4 + 90 (2x) ^ 3 - 270 (2x) ^ 2 + 405 (2x) - 243

B) (2x) ^ 5 + 15 (2x) ^ 4 - 90 (2x) ^ 3 + 270 (2x) ^ 2 - 405 (2x) + 243

C) (2x) ^ 5 + 15 (2x) ^ 4 + 90 (2x) ^ 3 + 270 (2x) ^ 2 + 405 (2x) + 243

D) 2 (x) ^ 5 - 30 (x) ^ 4 + 180 (x) ^ 3 - 540 (2x) ^ 2 + 810 (x) - 243

the answer is C

C

Step-by-step explanation:

(2x + 3) ^5 = C (5,0) 2x^5*3^0 +

C (5,1) 2x^4*3^1 + C (5,2) 2x^3*3^2 + C (5,3) 2x^2*3^3 + C (5,4) 2x^1*3^4 + C (5,5) 2x^0*3^5

Recall that

C (n, r) = n! / (n-r) ! r!

C (5,0) = 1

C (5,1) = 5

C (5,2) = 10

C (5,3) = 10

C (5,4) = 5

C (5,5) = 1

= 1 (2x^5) 1 + 5 (2x^4) 3 + 10 (2x^3) 3^2 + 10 (2x^2) 3^3 + 5 (2x^1) 3^4 + 1 (2x^0) 3^5

= 2x^5 + 15 (2x^4) + 90 (2x^3) + 270 (2x^2) + 405 (2x) + 243

= 32x^5 + 15 (16x^4) + 90 (8x^3) + 270 (4x^2) + 810x + 243

= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243