Question

Solve and write solution intervals

Answer 1

\[\frac{ x^3 - 2x^2 - 3x }{ x + 1 } \ge 0\]

Answer 2

Okay, so basically for this, you can do polynomial division.

Answer 3

You know how to do that right?

Answer 4

let me check and see if I remember

Answer 5

sorry it puts them a bit weird

Answer 6

So what is the resultant polynomial expression?

Answer 7

x^2 -3x ?

Answer 8

Yes, write it as \(x(x - 3)\ge0\)

Answer 9

hm ok

Answer 10

So basically, at this point you can either graph that on paper or you can make intervals based on critical points. Do you know the critical points?

Answer 11

It's much easier to just graph on paper btw. We know the quadratic expression is concave up because the leading term is positive.

Answer 12

I think my teacher wants me to use interval notation

Answer 13

You'll get the interval solution no matter which method you use.

Answer 14

Yeah but in class we did like the union shape

Answer 15

That's after you actually find the intervals. We have to find the intervals first.

Answer 16

oh

Answer 17

how do we do that

Answer 18

Do you not know know how to graph the expression on paper? If you can't graph it then the next step is to make a number line and plot the critical points.

Answer 19

I think we did it using the number line

Answer 20

Okay, so what are the critical points?

Answer 21

I am not sure

Answer 22

The critical points are the all the values of \(x\) that will make the expression on the left side equal zero.

Answer 23

0 and 3?

Answer 24

Correct. Now make your number line using the drawing pad

Answer 25

Highlight the critical points

Answer 26

So basically what you have just done is create three intervals:
\((-\infty, 0], [0,3], [3, \infty)\)


We have to figure out which of those intervals will be greater than or equal to zero.

Answer 27

I think I just remembered how to do this. I got (- infinity,0] and [3,infinity) as solutions

Answer 28

If we find a value along an interval that produces a result that is greater than zero, then we put a PLUS above it. If we find a value along the interval that produces a result that is less than zero then we put a MINUS above it.

Answer 29

That is correct.

Answer 30

hmm, was there any tricks to this one like the numerator/denominator thing a couple problems ago?

Answer 31

That wasn't a trick. That's just what I do anytime I see a rational expression.

Answer 32

What they teach in school about how to do these is silly.

Answer 33

Wait, what numerator/denominator thing? I thought you were talking about this problem at first.

Answer 34

Where if the numerator degree is larger than the denominator degree, there is no horizontal asymptote. Teachers like those things on tests so I was just wondering if there was anything like that on this one.

Answer 35

What I just showed you IS the trick. Trust me.

Answer 36

I don't know how your teacher showed you how to do these though. I mean you could try factoring, but eventually, she will give you one that won't factor so easily. Then you'll be stuck. The method I showed you works for just about any problem your teacher will give you on these if it involves a rational expression.

Answer 37

That's why I showed it to you. We could have just factored and got the same expression

Answer 38

oh wow you're right, just tried it

Answer 39

Go ahead and post your next one.

Answer 40

sorry I lost connection

Answer 41

Yeah, Server related crashes. Go ahead and post your next question.