What is #((3x^2 - 7x - 6) / (x^3 - 3x^2+4s-12)) /( (3x^2 - 4x - 4)/(x^4 - 16))#?

1 Answer
Nghi N.
Jul 26, 2016

(x + 2)

Explanation:

#f(x) = (a/b)/(c/d)#

Factor all the functions of x:
1. #a = 3x^2 - 7x -6 =# (3x + 2)(x - 3)
2. #b = x^3 - 3x^2 + 4x - 12 = (x - 3)(x^2 + 4)#
3. #c = 3x^2 - 4x - 4 =# (3x + 2)(x - 2)
4. #d = x^4 - 16 = (x^2 + 4)(x - 2)(x + 2)#

#f(x) = [(3x + 2)(x - 3)]/[(x - 3)(x^2 + 4)][(x^2 + 4)(x - 2)(x + 2)]/[(3x + 2)(x - 2)]#

f(x) = (x + 2)

Related questions
Impact of this question
434 views around the world
You can reuse this answer
Creative Commons License