Solving inequality for a polynomial

vrfulgen

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Jan 22, 2013
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So I just finished factoring a polynomial and I got:

p(x)=(x-2)(x^6-2x^5+3x^4-6x^3)


But now I have to solve the inequalities for this polynomial for p(x)>0 and p(x)<0.


I understand I have to factor it, and I got x^3(x^2+3)(x-2)^2

And then I would take the (x^2+3) and throw in some square roots and i's.



[FONT=arial, helvetica, clean, sans-serif]and I'll probably use a number line or something, I just don't know if I include the complex result of (x^2+3)?
I have to show my work so no calculator....

I'd appreciate any help. Thanks. :)
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So I just finished factoring a polynomial and I got:

p(x)=(x-2)(x^6-2x^5+3x^4-6x^3)


But now I have to solve the inequalities for this polynomial for p(x)>0 and p(x)<0.


I understand I have to factor it, and I got x^3(x^2+3)(x-2)^2

And then I would take the (x^2+3) and throw in some square roots and i's.



and I'll probably use a number line or something, I just don't know if I include the complex result of (x^2+3)?
I have to show my work so no calculator....

I'd appreciate any help. Thanks. :)
Provided x is a real number, x^2 + 3 is a real number. Furthermore, the complex numbers do not have order in the sense that you are used to. The problem is asking about ordering real values of p(x).

p(x)=(x2)(x62x5+3x46x3)=x3(x2)(x32x2+3x6)=(x0)3(x2)2(x2+3).\displaystyle p(x) = (x - 2)(x^6 - 2x^5 + 3x^4 - 6x^3) = x^3(x - 2)(x^3 - 2x^2 + 3x - 6) = (x - 0)^3(x - 2)^2(x^2 + 3).

Good job factoring. Now, within the domain of the real numbers, under what circumstances can (x2)2<0?\displaystyle (x - 2)^2 < 0?

Still staying within the domain of the real numbers, under what circumstances can (x2+3)<0?\displaystyle (x^2 + 3) < 0?

So under what circumstances can p(x) > 0? Under what circumstances can p(x) < 0?
 
So I just finished factoring a polynomial and I got:

p(x)=(x-2)(x^6-2x^5+3x^4-6x^3)

But now I have to solve the inequalities for this polynomial for p(x)>0 and p(x)<0.


I understand I have to factor it, and I got x^3(x^2+3)(x-2)^2

And then I would take the (x^2+3) and throw in some square roots and i's.



and I'll probably use a number line or something, I just don't know if I include the complex result of (x^2+3)?
I have to show my work so no calculator....

I'd appreciate any help. Thanks. :)
You don't include complex numbers on real number line! x2+3\displaystyle x^2+ 3 has no real solutions because it positive for all x. Since "positive times positive is positive" and "positive times negative is negative", a factor that is always positive can be ignored. The first factor, x3\displaystyle x^3 is 0 only when x= 0 and the last, (x2)2\displaystyle (x- 2)^2 is 0 only at x= 2.

So we need to look at three intervals:
x< 0. Now, x3\displaystyle x^3 is negative while both x2+3\displaystyle x^2+ 3 and (x2)2\displaystyle (x- 2)^2 are positive. The product "(-)(+)(+)" is "-" so p(x)< 0 for all x< 0.

0< x< 2. Now x3\displaystyle x^3 is positive while both x2+3\displaystyle x^2+ 3 and (x2)2\displaystyle (x- 2)^2 are positive. The product "(+)(+)(+)" is "+" so p(x)> 0 for all 0< x< 2.

x> 2. Clearly the same thing happens- the only reason I mention x=2 is that p(2)= 0 so neither p(x)> 0 or p(x)< 0.

So p(x)> 0 for 0< x< 2 and x> 2. p(x)< 0 for x< 0.
 
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