The binomial theorem states, (x+a)^n=sum_"k=0"^n((n!)/(k!(n-k)!))x^"n-k"a^k
Here, n=5, x=2x,a=-3
(2x-3)^5=sum_"k=0"^5((5!)/(k!(5-k)!))(2x)^"5-k"(-3)^k
When k=0,
((5!)/(0!(5-0)!))(2x)^"5-0"(-3)^0
(120/(1*120))*32x^5*1
=32x^5
When k=1,
((5!)/(1!(5-1)!))(2x)^"5-1"(-3)^1
120/(1*24)*16x^4*(-3)
=-240x^4
When k=2,
((5!)/(2!(5-2)!))(2x)^"5-2"(-3)^2
120/(2*6)*8x^3*9
=720x^3
When k=3,
((5!)/(3!(5-3)!))(2x)^"5-3"(-3)^3
120/(6*2)*4x^2*(-27)
=-1080x^2
When k=4,
((5!)/(4!(5-4)!))(2x)^"5-4"(-3)^4
120/(24*1)*2x*81
=810x
When k=5,
((5!)/(5!(5-5)!))(2x)^"5-5"(-3)^5
120/(120*1)*1*(-243)
=-243
Add each individual answer up. The answer is 32x^5-240x^4+720x^3-1080x^2+810x-243.
