x+2)^3(X-1)^4(x+3)^2 <0
Write this solution in interval notation.
Can someone explain how to do the method where you start with a + on the number line, and depending on the odd or even value of the next exponent, you switch to and -/+ ?
I do these in the following way.
If this had been an equation, the critical values, or the x-intercepts would have been
x = -2,1, and -3
splitting the number line into 4 segments
I then take any value in each of those sections and see if it satisfies the inequation
We don't need the actual answer just the signs
1. x < -3, say x = -5
(-)(+)(+) = (-) which is < 0 , good
2. x between -3 and -2, say -2.5
(-)(+)(+) = (-) which is < 0, good
3. between -2 and 1 , say x=0
(+)(+)(+) > 0 NO good
4. x > 1 ,say x=5
(+)(+)(+) > 0 , NO good
so x < -2 , x ≠ -3
In this case we could have seen the solution in a rather straightforward case.
Since the last two factors have even exponents, those last two factors have to be ≥ 0, so the negative could only come from the first factor of
(x+2)^3 , which of course is true for x < -2
but that would include the case of x = -3 which would make the last factor equal to zero, thus the whole thing would be zero, and not < 0
That is why I had to exclude x = -3 from my interval.