How do I find the binomial expansion of #(3x-2)^4#?

1 Answer
AJ Speller
Sep 28, 2014

Note that the notation for the binomial coefficients is incorrect . It should not look like a fraction. The fraction bar should be removed the numbers should be on top of each other vertically.

#=(4/0)(3x)^4(-2)^0+(4/1)(3x)^3(-2)^1+(4/2)(3x)^2(-2)^2+(4/3)(3x)^1(-2)^3+(4/4)(3x)^0(-2)^4#

#=(1)(3x)^4(-2)^0+(4)(3x)^3(-2)^1+(6)(3x)^2(-2)^2+(4)(3x)^1(-2)^3+(1)(3x)^0(-2)^4#

#=(3x)^4+(4)(3x)^3(-2)+(6)(3x)^2(-2)^2+(4)(3x)(-2)^3+(-2)^4#

#=81x^4-216x^3+216x^2+96x+16#

Related questions
See all questions in Pascal's Triangle and Binomial Expansion
Impact of this question
12979 views around the world
You can reuse this answer
Creative Commons License