How do you use the Binomial Theorem to expand #(x-3)^5#?

1 Answer
Jun 30, 2016

Answer:

#(x-3)^5=x^5-15x^4+90x^3-270x^2+405x-243.#

Explanation:

The Binomial Theorem Expansion #: (x+y)^n=x^n+""nC_1x^(n-1)y+""nC_2x^(n-2)y^2+""nC_3x^(n-3)y^3+.........+""nC_rx^(n-r)y^r+... ...+y^n. #

Letting #y=-3, n=5#, we get,
#(x-3)^5=x^5+""5C_1x^4(-3)^1+""5C_2x^3(-3)^2+""5C_3x^2(-3)^3+""5C_4x^1(-3)^4+(-3)^5.#

Here, #""5C_1=""5C_(5-1)=""5C_4=5, ""5C_2=""5C_3=(5*4)/(1*2)=10#

So, #(x-3)^5=x^5+5x^4(-3)+10x^3(9)+10x^2(-27)+5x(81)+(-243).#
#=x^5-15x^4+90x^3-270x^2+405x-243.#