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slamgirl [31]
3 years ago
14

What is the binomial expansion of (2x – 3)5?

Mathematics
2 answers:
Sveta_85 [38]3 years ago
8 0
The answer is A............
____ [38]3 years ago
4 0

<u>Answer-</u>

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

<u>Solution-</u>

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


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