Answer:
wait who is that again?
Step-by-step explanation:
Answer:
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
No its a negitive number one x minus five x's equal -14
