1. ## Inequality help

So I have the following question:

(x+3)^5 (x-1)(x-4)^2 < 0

I know the roots are at -3, 1, 4 and the the intervals are:
(-infinite, -3)
(-3, 1)
(1, 4)
(4, infinity)

However, I don't know how to determine if an interval is positive or negative. Can someone help me? Thanks.

2. Put the roots on line(x-axis), check if an interval is positive or negative by substituting i.e: x=-4 {(-infinite, -3)} to (x+3)^5 (x-1)(x-4)^2,
if (x+3)^5 (x-1)(x-4)^2 positive the sing is + else -

At the end of process you will get your roots on line(x-axis)and sing in witch interval.

(x+3)^5 (x-1)(x-4)^2 < 0

so you take all intervals with sing minus.

3. Oh yeah! How could I be so silly!

Thank you very much!

4. Another way to do this is to choose one point in each interval and calculate the value for that point. If the value is -, then it is < 0 for the entire interval.

5. Originally Posted by HallsofIvy
Another way to do this is to choose one point in each interval and calculate the value for that point. If the value is -, then it is < 0 for the entire interval.
This is the method I've been using. Thanks!

6. Originally Posted by Glitch
So I have the following question:

(x+3)^5 (x-1)(x-4)^2 < 0
$(x+3)^5(x-1)(x-4)^2=(x+3)^4(x-4)^2(x+3)(x-1)<0,$ so we only need that $(x+3)(x-1)<0$ which is very easy to solve since $x^2+2x-3=(x+1)^2-4<0$ thus $-2 is the solution set.