\[\dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0\]
Find all the possible values of \(x\).
+118609
\(\dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0\)
The first thing I notice is the 1st and last term on the bottom are cubic.
so I check but the long bracket simplifies to (x+2)^3.
The top also looks promising ... (2x+3)^3
so we have:
\(\dfrac{(2x+3)^3{(x-1)}^2}{(x+2)^3{(x+1)}^2} < 0\)
x cannot be -2 or -1 because I can't divide by 0
x is not going to be -1.5 or 1 either since 0 is not less than 0
if x is not 1 or -1 then both (x-1)^2 and (x+1)^2 will be positive
cubics have the same sign as the term that is being cubed so I only need to think about
2x+3 and x+2
If they have the same sign, both pos or both neg then the outcome will be positive
If they have different signs the outcome will be negative.
So we want
2x+3<0 and x+2>0 OR 2x+3>0 and x+2<0
x < -1.5 and x > -2 OR x > -1.5 and x <-2
-2 < x < -1.5 OR no solution
so
-2 < x < -1.5
Here is the graph
