polynomial inequalities

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mathshelpplease

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[math](x+3)(x^2-4)≤(x^2+x-6)[/math]
Hello, I'm struggling with this problem. Any idea? I think I may have got my division wrong as I only find one solution where I know there should be more.
 
mathshelpplease said:
[math](x+3)(x^2-4)≤(x^2+x-6)[/math]
Hello, I'm struggling with this problem. Any idea? I think I may have got my division wrong as I only find one solution where I know there should be more.
Click to expand...

Don't (try to) divide, please! You lost information.

Factor after some manipulation.

I assume you factored or had it in mind:

\(\displaystyle (x + 3)(x - 2)(x + 2) \ \le \ (x + 3)(x - 2)\)

\(\displaystyle (x + 3)(x - 2)(x + 2) - (x + 3)(x - 2) \ \le \ 0\)

\(\displaystyle (x + 3)(x - 2)[x + 2 - 1] \ \le \ 0\)

\(\displaystyle (x + 3)(x - 2)(x + 1) \ \le \ 0\)


Can you please take it to the end by doing the usual region testing? Please show your work here if your answer(s)
do not agree with an answer source (if you have access to one).
 
Last edited:
When you divide an inequality by a negative value you need to change the direction of the inequalities.

So if you divide by sides of an inequality, for example, by x-9 then do you change the direction of the inequality?

Well if x<9 (which means that x-9 is negative), then you must change the direction of the inequality sign. However if x > 9 (then x-9>0) you will not need to chance the sign of the inequality. If x=9 (x-9=0), then you are dividing by 0 which is not allowed.

Now you can do this in cases where one case is x<9 and another case x>9 but it can be complicated especially when you can do it like pka suggested and not have cases to consider.
 
lookagain said:
Don't (try to) divide, please! You lost information.

Factor after some manipulation.

I assume you factored or had it in mind:

\(\displaystyle (x + 3)(x - 2)(x + 2) \ \le \ (x + 3)(x - 2)\)

\(\displaystyle (x + 3)(x - 2)(x + 2) - (x + 3)(x - 2) \ \le \ 0\)

\(\displaystyle (x + 3)(x - 2)[x + 2 - 1] \ \le \ 0\)

\(\displaystyle (x + 3)(x - 2)(x + 1) \ \le \ 0\)


Can you please take it to the end by doing the usual region testing? Please show your work here if your answer(s)
do not agree with an answer source (if you have access to one).
Click to expand...
would the values of x that satisfy this equation then be
[math]x≤−3,x≤2,x≤−1[/math]
thank you for your help!
 
By x<=-3, x<=2, x<-1 do you mean x<=-3 AND x<=2 AND x<=-1 or do you mean x<=-3 OR x<=2 OR x<=-1.

You do realize that the two results are not the same.
 
mathshelpplease said:
would the values of x that satisfy this equation then be
[math]x≤−3,x≤2,x≤−1[/math]
Click to expand...

It is not an equation. It is an inequality.

I mentioned region testing. When you set those three binomial factors equal
to zero, you get, when you arrange them from smallest to largest, the critical
numbers of x = -3, -1, 2.

Those three critical numbers correspond to four regions:

(-oo, -3], [-3, -1], [-1, 2], and [2, oo).

Pick convenient test x-values from each region to substitute into the
inequality (the last form of it I wrote will do) to see which region(s) to
include as part of your answer. If when you substitute a test x-value,
and it satisfies the inequality, then you include that particular region
that that x-value lies in.

Suggested test x-values here might be: -4, -2, 0, and 3, respectively.
 
lookagain said:
It is not an equation. It is an inequality.

I mentioned region testing. When you set those three binomial factors equal
to zero, you get, when you arrange them from smallest to largest, the critical
numbers of x = -3, -1, 2.

Those three critical numbers correspond to four regions:

(-oo, -3], [-3, -1], [-1, 2], and [2, oo).

Pick convenient test x-values from each region to substitute into the
inequality (the last form of it I wrote will do) to see which region(s) to
include as part of your answer. If when you substitute a test x-value,
and it satisfies the inequality, then you include that particular region
that that x-value lies in.

Suggested test x-values here might be: -4, -2, 0, and 3, respectively.
Click to expand...
thank you,

following on from this I've found
[math]-3≤x≤-1, 2≤x[/math]
to be two regions which satisfy this inequality - would this be the final answer?
 
mathshelpplease said:
thank you,

following on from this I've found
[math]-3≤x≤-1, 2≤x[/math]
to be two regions which satisfy this inequality - would this be the final answer?
Click to expand...

I have been away for several hours. Those intervals do not work. If you were
to check test numbers of x = -2 and x = 3, for example, in my last inequality ,
of post #2, you would see that they do not satisfy it.

Look at the first test value of x = -4 suggested in post #8. It will be substituted
into three places in the inequality. If it satisfies, then include that region, else
discard that region. And so on.
 
Screenshot 2021-10-08 at 08.58.58.png

Sorry I may be completely mistaken, but if this is the plotted graph are the regions not correct? I am very new to this topic sorry and genuinely unsure how to proceed
 
mathshelpplease said:
View attachment 29159

Sorry I may be completely mistaken, but if this is the plotted graph are the regions not correct? I am very new to this topic sorry and genuinely unsure how to proceed
Click to expand...
It appears that your solution was the set of values where this (the black curve) is positive or zero; what you want is the set where it is negative or zero, in order to satisfy the original inequality. You just got something backward.

I don't know what the other lines represent.
 
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