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#.
