This is a nasty one. so hang on...
Solving quadratic or higher degree inequalities is similar to solving like equations. We want one side zero and the other side factored. You already have one side zero so we start with factoring.
There is no GCF (other than 1). The expression has too many terms for any of the patterns or for trinomial factoring. And I don't see a way to use factoring by grouping so we're left with factoring by trial and error of the possible rational roots.
The possible rational roots of are all the positive and negative numbers which can be formed using a factor of 30 over a factor of 2. The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30 and the factors of 2 are 1 and 2. This makes for a lot of possible roots. It could take a while to find one. I'll save you some time and show one. The simplest way to check a possible root is to use synthetic division:
3 | 2 3 -17 -30
--- 6 27 30
-------------------
2 9 10 0
Since the remainder is zero then 3 is a root. And if 3 is a root, then (x-3) is a factor. We can get the other factor from the division:
The trinomial factor will factor, too, so now we have:
So the factored inequality is:
which tells that the product of the three factors is negative. And how can the product of three numbers turn out negative? You should understand that this can happen only if "all three factors are negative" or "two factors are positive and one is negative". Any other combination of factors will result in positive or zero products. So we just have to express "all three factors are negative" or "two factors are positive and one is negative" in the form of inequalities and then solve those inequalities.
For "all three factors are negative":
x-3 < 0 and 2x+5 < 0 and x+2 < 0
For "two factors are positive and one negative" we have to account for all possible combinations of two positive and one negative factor:
((x-3 < 0 and 2x+5 > 0 and x+2 > 0) or (x-3 > 0 and 2x+5 < 0 and x+2 > 0) or (x-3 > 0 and 2x+5 > 0 and x+2 < 0))
So the solution comes from the combination of these two:
(x-3 < 0 and 2x+5 < 0 and x+2 < 0) or ((x-3 < 0 and 2x+5 > 0 and x+2 > 0) or (x-3 > 0 and 2x+5 < 0 and x+2 > 0) or (x-3 > 0 and 2x+5 > 0 and x+2 < 0))
If we were lucky we would have factors whose order (largest, smallest and in between) we can determine. But with the (2x+5) factor we are not lucky. (If you want to see a problem where the order can be determined and how that makes the problem much easier, click here.)
To solve this problem we just solve each individual simple inequality, making sure to keep track of the parentheses, the and's and the or's.
(x < 3 and x < -5/2 and x < -2) or ((x < 3 and x > -5/2 and x > -2) or (x > 3 and x < -5/2 and x > -2) or (x > 3 and x > -5/2 and x < -2))
Now we have several expressions in parentheses with three inequalities joined by and's. These expressions can be reduced.
(x < 3 and x < -5/2 and x < -2)
can be reduced to:
(x < -5/2)
After all, if a number is less than -5/2, won't it also be less than 3 and -2?
(x < 3 and x > -5/2 and x > -2)
can be reduced to
(x < 3 and x > -2)
After all, if a number is greater than -2, won't it also be greater than -5/2?
I hope you are seeing how the "and" of two greater than's or the "and" of two less than's can be reduced to just one of the two (because I am going to jump to the reduced solution:
(x < 3 and x < -5/2 and x < -2) or ((x < 3 and x > -5/2 and x > -2) or (x > 3 and x < -5/2 and x > -2) or (x > 3 and x > -5/2 and x < -2))
reduces to
(x < -5/2) or ((x < 3 and x > -2) or (x < -5/2 and x > 3) or (x < -2 and x > 3))
We can do even more simplifying. Look at the last two parenthesized expressions:
(x < -5/2 and x > 3)
(x < -2 and x > 3)
With an "and" both inequalities must be true. But how do you get a number that is less than -5/2 and greater than 3 at the same time? And for the second expression, how is a number less than -2 and greater than 3 at the same time? The answer for both: it's not possible. So there will be no solutions coming from these two. So our solution is now:
(x < -5/2) or (x < 3 and x > -2)
which tells us that any number less than -5/2 works. And any number between 3 and -2 will work also.
So this was a challenger because:- The factoring was not simple.
- Once factored the factors could not be ordered. When you can't order the factors the inequalities we need are harder to write, harder to solve and harder to reduce and simplify.