# Help polynomial

1. Apr 1, 2004

### 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

2. Apr 1, 2004

### 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)...

3. Apr 1, 2004

### expscv

this has complexy number i involved~

4. Apr 1, 2004

### expscv

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

Last edited: Apr 1, 2004
5. Apr 1, 2004

### 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).

6. Apr 1, 2004

### rattis

did you post this in maths section as well?