# Another problem, almost done :)

• Mar 19th 2006, 11:59 PM
kharris82
Another problem, almost done :)
Solve: 1/(x-2) less then or equal to 0

Also having trouble with: "Suppose that a polynomial function of degree 4 with rational coefficients has -4, -5, -4, -i as zeros. Find the other zeros.
A)4,5,-4,i
B)4 + i
C)-4 + i
D) -4, -i

Thank you captain black!
• Mar 20th 2006, 10:25 AM
earboth
Quote:

Originally Posted by kharris82
Solve: 1/(x-2) less then or equal to 0

Also having trouble with: "Suppose that a polynomial function of degree 4 with rational coefficients has -4, -5, -4, -i as zeros. Find the other zeros.
A)4,5,-4,i
B)4 + i
C)-4 + i
D) -4, -i

Thank you captain black!

Hello,

I can offer you some help with your first problem. But I don't know what to do with your 2nd problem: The given list of zeros looks a little bit funny to me (for instance why is the (-4) listed twice?).

to 1.: You're looking for negative quotient with a positive numerator. That is only possible if the nominator is negative:
$\frac{1}{x-2}\leq 0\ \Longrightarrow \ x-2<0 \ \Longrightarrow \ x<2$

Greetings

EB
• Mar 20th 2006, 03:10 PM
topsquark
Quote:

Originally Posted by kharris82
Solve: 1/(x-2) less then or equal to 0

Also having trouble with: "Suppose that a polynomial function of degree 4 with rational coefficients has -4, -5, -4, -i as zeros. Find the other zeros.
A)4,5,-4,i
B)4 + i
C)-4 + i
D) -4, -i

Thank you captain black!

As to the second problem, if the polynomial (with rational coefficients) is of degree 4 then it has at most 4 distinct roots. In addition, if a complex root of a polynomial function with rational coefficients exists, then the complex conjugate of that root is also a root. For that reason, "i" must also be a root of this polynomial. But that means your polynomial is of degree 5, not 4.

I don't understand what they mean by "find the other zeros." You've already got four listed!

Something's fishy here.

-Dan