Solving inequality for a polynomial

[vrfulgen]

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. [/FONT]

 

[JeffM]

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).

\(\displaystyle p(x) = (x - 2)(x^6 - 2x^5 + 3x^4 - 6x^3) = x^3(x - 2)(x^3 - 2x^2 + 3x - 6) = (x - 0)^3(x - 2)^2(x^2 + 3).\)

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

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

So under what circumstances can p(x) > 0? Under what circumstances can p(x) < 0?

 

[HallsofIvy]

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! \(\displaystyle x^2+ 3\) 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, \(\displaystyle x^3\) is 0 only when x= 0 and the last, \(\displaystyle (x- 2)^2\) is 0 only at x= 2.

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

0< x< 2. Now \(\displaystyle x^3\) is positive while both \(\displaystyle x^2+ 3\) and \(\displaystyle (x- 2)^2\) 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.

 

[vrfulgen]

THANK YOU so much!!