help polynomial

[expscv]

if p(x ) = x^6 + x^4 + x^2 + 1

show that the solutions of the equation p(x ) = 0 are among the solutions of the equation x^8 - 1 = 0

hence factorse p(x ) fully over R

[Chen]

Excuse my ignorance but I don't see how p(x) can ever intersect with the X axis? Its minimum point is (0, 1)... :confused:

[expscv]

this has complexy number i involved~

[expscv]

answer = (x^2+1)(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1) but i dont see how

[Chen]

p_{(x)} = x^6 + x^4 + x^2 + 1 = x^4(x^2 + 1) + x^2 + 1 = (x^2 + 1)(x^4 + 1) = 0

So either (1) x^4 = -1 or (2) x^2 = -1

Now what are the solutions for x^8 = 1?
x^4 = \pm 1
If x^4 = -1 we see that it will include the two solutions of equation (1). If x^4 = 1 then we can say that x^2 = \pm 1 and again, if x^2 = -1 we see it will include the two solutions of equation (2).

[rattis]

did you post this in maths section as well?