Question

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

Answer 1

the answer is C

Explanation:

Answer 2

C

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