Answer-
The binomial expansion of the given function is,
[tex]\Rightarrow (2x-3)^5=(2x)^5-15(2x)^{4}+90(2x)^{3}-270(2x)^{2}+405(2x)-243[/tex]
Solution-
The general form of Binomial Expansion is,
[tex](x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+......+\binom{n}{n}x^0y^n[/tex]
Putting x = 2x, y = -3, and n = 5
[tex](2x+(-3))^5=\binom{5}{0}(2x)^5(-3)^0+\binom{5}{1}(2x)^{5-1}(-3)^1+\binom{5}{2}(2x)^{5-2}(-3)^2+\binom{5}{3}(2x)^{5-3}(-3)^3+\binom{5}{4}(2x)^{5-4}(-3)^4+\binom{5}{5}(2x)^{5-5}(-3)^5[/tex]
[tex]\Rightarrow (2x-3)^5=1.(2x)^5.1\ \ +\ \ 5.(2x)^{4}.(-3)\ \ +\ \ 10.(2x)^{3}.(9)\ \ +\ \ 10.(2x)^{2}.(-27)\ \ +\ \ 5.(2x)^{1}.(81)\ \ +\ \ 1.(2x)^{0}.(-243)[/tex]
[tex]\Rightarrow (2x-3)^5=(2x)^5-15(2x)^{4}+90(2x)^{3}-270(2x)^{2}+405(2x)-243[/tex]
