Question

Suppose that R(x) is a polynomial of degree 7 whose coefficients are real numbers.
Also, suppose that R(x) has the following zeros.
-2+i, 2i

Answer the following.
(a) Find another zero of R(x) .
(b) What is the maximum number of REAL zeros that R(x) can have?
(c) What is the maximum number of NONREAL zeros that R(x) can have?

Answer 1

NEED TO KNOW IF THIS IS CORRECT GUYS- PLEASE ANYONE...

A. From the Complex root theorem, imaginary numbers come in pairs, so another zero to -2i+i is -2-i, and another zero to 2i is -2i

B. The Fundamental theorem of Algebra basically says to whatever degree the polynomial is, there will be that many zeroes. So far we have accounted for 4 zeroes: -2+i, -2-i, 2i, and -2i. ZERO of them real, 4 complex. So if we subtract 7 from 4 , we have 3 more zeroes. So in total there could be a max of 3 + 3=6 real zeroes

C. The Complex root theorem but basically says that nonreal zeroes can only exists as pairs so 7- 4= 3 two of those could be nonreal, so there could only be a max of 4 non real zeroes.

Answer 2

A. IF you have Real Coefficients.

B. Seven (7) is the complex total. We have accounted for 4. This leaves 3 possibel Real. It is odd degree, so there must be one (1) Real. There may be 3.

C. I don't understand your logic. We have 4 complex already identified. We must have one Real. This leaves 2. We know nothing of these. The could be Complex Conjugates or they could be Real.

Answer 3

tk for A. they are are real coefficients.

for B. are you saying that there is only 1 real #?

for C. are you saying that the maximum number of nonreal zeros is 2 or 4?

Answer 4

@satellite73 help me please!!

Answer 5

A: True, but you didn't state that or otherwise point it out.
B: 1 or 3
C: 6

The only possibilities are

Complex: 4 Real: 3 Total: 7
Complex: 6 Real: 1 Total: 7