Polynomials Question (1 Viewer)

[Web Addict]

Hi.

P(x)=16x^4-32x^3+16x^2+kx-5, where k is an integer. P(x) has two rational roots, which are opposites of each other, and two non-real roots.
a. If a (alpha) is a non-real root of p(x), show that Re(a)=1 and modulus(a)>1.
b. If the rational roots are +- b (beta), deduce that b^2<5/16/
c. Find the rational roots and the value of k.
d. Factor P(x) into irreducible factors with integer co-efficients.

I just need help with part c. I've done part a and part b.

Thank you.

 

[erckle999]

By two rational roots that are opposites of each other do you mean they are the negative of each other or that they are reciprocals of each other?

 

[Web Addict]

I think its means that one is positive and one is negative. It works for part a and part b if this is the case.

 

[Exia_]

sub x = b and x=-b into the polynomial to obtain 2 equations.
16b^4 -32b^3 +16b^2 +kb -5 = 0
16b^4 +32b^3 +16b^2 -kb -5 = 0
You can then add these 2 equations and find b. (b = +-1/2)
Once you find b, sub it back into the polynomial to find k.

Btw, is this a troung student since this question was in our homework this week.

 

[Web Addict]

Thanks and yeah, I am a Troung student.

 

[kev-]

For b), I got b^2 = 5/16. how do you introduce the inequality?
any help is appreciated

 

[such_such]

For b), I got b^2 = 5/16. how do you introduce the inequality?
any help is appreciated

Um, for (b) I would do product of roots first:
(-b^2) |a|^2 = -5/16
(b^2) |a^2| = 5/16

|a|>1 (you proved that in a)
|a|^2 > 1
b^2|a|^2 > b^2
5/16 > b^2 (from your product of roots you know that (b^2) |a^2| = 5/16 )

therefore b^2 < 5/16