How do you use the Binomial Theorem to expand #(3x - 2y^2)^5#?

1 Answer

Answer:

#(3x-2y^2)^5=#
#243x^5-810x^4y^2+1080x^3y^4-720x^2y^6+240xy^8-32y^10#

Explanation:

The formula of Binomial Theorem

#(a+b)^n=((n),(0))a^n b^0+((n),(1))a^(n-1) b^1+((n),(2))a^(n-2) b^2+...+#
#((n),(n-1))a^1b^(n-1)+((n),(n))a^0b^(n)#

Let #a=3x# and #b=-2y^2#

#(3x-2y^2)^5=#

#(3x)^5(-2y^2)^0+5/(1!)(3x)^4(-2y^2)^1+(5*4)/(2!)(3x)^3(-2y^2)^2+(5*4*3)/(3!)(3x)^2(-2y^2)^3+(5*4*3*2)/(4!)(3x)^1(2y^2)^4+(5*4*3*2*1)/(5!)(3x)^0(-2y^2)^5#

#=243x^5-5(81)(2)x^4y^2+10(27)(4)x^3y^4+10(9)(-8)x^2y^6+5(3)(16)xy^8-32y^10#

#=243x^5-810x^4y^2+1080x^3y^4-720x^2y^6+240xy^8-32y^10#

God bless you...