# Simple question =/ (x-1)(x+2)(x-4) ≤ 0

1. Jan 25, 2009

### rought

1. The problem statement, all variables and given/known data

Solve: (x-1)(x+2)(x-4) ≤ 0

3. The attempt at a solution

I know how to do this if it's equal to zero, it would just be X=1,-2,4 but how do you do it with the less than or equal to thingy?

2. Jan 25, 2009

### epenguin

You don't know the answer or you don't know how to set it out formally?
If the first, draw a graph, and the proof will probably come to you.

e.g.
For the product of three things to be negative, how many of the things have to be negative?
For it to change signs, e.g. as x increases, what has to happen?

3. Jan 25, 2009

### Дьявол

You got 3 cases:

I case:

x-1 ≤ 0

x+2 > 0

x-4 > 0

II case

x-1 > 0

x+2 ≤ 0

x-4 > 0

III case:

x-1 > 0

x+2 > 0

x-4 ≤ 0

And then find union of all cases

4. Jan 25, 2009

### yeongil

One way is to make a sign chart. Draw a number line, marking the three zeros like this:
Code (Text):

0          0          0
----------+----------+----------+-----------
-2          1          4

(I wrote a "0" on top of the three numbers to indicate the zeros in the problem.)
You have four intervals to check: (-∞, -2), (-2, 1), (1, 4), and (4, ∞). Within each interval, indicate the sign of each factor of the inequality. Here is the sign chart again with the first interval done:
Code (Text):

(-)(-)(-) 0          0          0
----------+----------+----------+-----------
neg.  -2          1          4

For any number in (-∞, -2) each factor would be negative, so I would write a "(-)" 3 times. The product would be negative. (Indicate a positive factor with a "(+)".)

Now try it with the other three intervals. You'll find your answer.

P.S. OK, you don't really need to do this if you are aware as to how graphs of functions look. The left side of the inequality is a cubic and it crosses the x-axis three times. The leading coefficient (if you multiplied the factors out) is positive, so that should tell you something about the end behavior of this polynomial.

01