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

d) (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 A............

Answer 2

Answer-

The binomial expansion of the given function is,

$\Rightarrow (2x-3)^5=(2x)^5-15(2x)^{4}+90(2x)^{3}-270(2x)^{2}+405(2x)-243$

Solution-

The general form of Binomial Expansion is,

$(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+......+\binom{n}{n}x^0y^n$

Putting x = 2x, y = -3, and n = 5

$(2x+(-3))^5=\binom{5}{0}(2x)^5(-3)^0+\binom{5}{1}(2x)^{5-1}(-3)^1+\binom{5}{2}(2x)^{5-2}(-3)^2+\binom{5}{3}(2x)^{5-3}(-3)^3+\binom{5}{4}(2x)^{5-4}(-3)^4+\binom{5}{5}(2x)^{5-5}(-3)^5$

$\Rightarrow (2x-3)^5=1.(2x)^5.1\ \ +\ \ 5.(2x)^{4}.(-3)\ \ +\ \ 10.(2x)^{3}.(9)\ \ +\ \ 10.(2x)^{2}.(-27)\ \ +\ \ 5.(2x)^{1}.(81)\ \ +\ \ 1.(2x)^{0}.(-243)$

$\Rightarrow (2x-3)^5=(2x)^5-15(2x)^{4}+90(2x)^{3}-270(2x)^{2}+405(2x)-243$