How do you solve #(x-3)(x-4) / (x-5)(x-6)^2<0#?
1 Answer
Feb 9, 2018
Explanation:
The polynomial has 4 roots:3,4,5 and 6.
Each term will be negative if x<root
equal to zero if x=root
Bigger than zero if x>root
A table of + and - would be almost impossible to do here, so let's make a list:
if x<3: all terms are negative so the multiplication gives negative
if x=3 polynomial equal to 0
if x in ]3,4[, (x-3) is positive and (x-4) (x-5) are negative, so the multiplication will give positive
if x=4 polynomial equal to 0
if x in ]4,5[, only (x-5) is negative so the multiplication will give negative
if x=5 polynomial is undefined
if x in ]5,+oo[, all terms are positive so multiplication will give positive
(x−6)^2 will always positive except when x=6.
The result is
