+0

help inequality

0
38
1

\[\dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0\]

Find all the possible values of \(x\).

Jan 13, 2021

1+0 Answers

#1
+112062
+1

\(\dfrac{(8x^3 + 36x^2 + 54x + 27){(x-1)}^2}{(x^3 + 6x^2 + 12x + 8){(x+1)}^2} < 0\)

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

Jan 13, 2021
edited by Melody  Jan 13, 2021