# Inequality help

• Jul 7th 2010, 11:21 PM
Glitch
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.
• Jul 8th 2010, 12:13 AM
Also sprach Zarathustra
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.
• Jul 8th 2010, 12:56 AM
Glitch
Oh yeah! How could I be so silly!

Thank you very much!
• Jul 8th 2010, 04:27 AM
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.
• Jul 8th 2010, 04:30 AM
Glitch
Quote:

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!
• Jul 8th 2010, 06:24 AM
Krizalid
Quote:

Originally Posted by Glitch
So I have the following question:

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

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