SOLUTION: please help me solve this inequality: (x+2)(x-1)(x-3)<= 0

Question 12596: please help me solve this inequality: (x+2)(x-1)(x-3)<= 0
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I've seen that you have been trying for a long time to get an answer to this one. I'm not sure the best way to explain it to you, since the graphing calculator is the BEST way to do it. Not knowing your background, I'll try a general explanation of sorts.

First solve the equation to find out where (x+2)(x-1)(x-3) = 0. This happens at three points on a numberline, at x= -2, x= 1, and at x=3. Now, the question is actually where will the product of these numbers be LESS THAN or EQUAL to ZERO. So you know THREE solutions right off the bat.

Now, you must realize that, since this is an INEQUALITY, it could be any of the intervals that are created by these three endpoints. If you graph the three points on a numberline, you will notice that these THREE numbers divide the numberline into FOUR intervals (always one more interval than endpoints!). What you need to do is to TEST out ONE POINT in each of the intervals.

Start with the values on the numberline to the RIGHT of x=3. For example, if you took 4 or any other number larger than x=3, and subsitute into the inequality: (x+2)(x-1)(x-3) <=0, it will be positive number times positive times a positive number, which is NOT less than zero.

Next, test a point that is between 1 and 3, like for example 2. If you subsitute x= 2, it comes out to be a product that IS less than zero.

Next, test a point that is between -2 and 1. That will be a positive, so it is NOT less than zero.

Finally, test a point that is less than -2, and the result will be a product that is NEGATIVE, so this interval works.

Therefore the answer is all values from negative infinity to -2 including the -2, and also from 1 to 3 including those endpoints.

It has occurred to me that what I just wrote is probably not a very clear explanation. Let's try the graphing method. You must graph y=(x+2)(x-1)(x-3), and find all values of x for which points on the graph are ON or BELOW the X-axis.

From this graph you can see that the graph is BELOW the x-axis from negative infinity to -2, and again from 1 to 3. Of course include the endpoints, but don't EVER include infinity.

R^2 at SCC

P.S. Do you see why no one wanted to answer this question???
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