Prove that the following polynomials has no integer roots

x^3+2x^2-x+5

The polynomial has no roots mod 2.

Thank you very much.

Results 1 to 5 of 5

- Jan 11th 2007, 12:37 PM #1

- Joined
- Nov 2006
- Posts
- 123

- Jan 11th 2007, 12:49 PM #2
In both cases the rational roots theorem works.

The above polynomial will only have rational roots if those rational roots are of the form:

$\displaystyle \frac{\text{factors of 5}}{\text{factors of 1}}$

To be precise the numerator is factors of the constant term, the denominator is factors of the coefficient of the leading term.

So in this case we only have four possible rational roots: $\displaystyle \pm1, \pm5$. Check these and you have your answer.

The same works for the polynomial mod(2):

$\displaystyle x^3+2x^2-x+5 \equiv x^3 + x + 1$ (mod 2)

Here we only have two possibles: $\displaystyle \pm 1$.

-Dan

- Jan 11th 2007, 12:58 PM #3

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Jan 11th 2007, 01:33 PM #4
Every cubic with real coefficients has at least 1 real solution.

The discriminant of a cubic is:

$\displaystyle 4b^{3}d-b^{2}c^{2}+4ac^{3}-18abcd+27a^{2}d^{2}$

If the discriminat is >0, then there is one real root and the other 2 are complex conjugates.

Checking this discriminant with a=1, b=2, c=-1, d=5, we get D=1007.

Therefore, there is one real root. The other 2 are complex, so they are eliminated as integer roots.

By the Intermediate Value Theorem, we find a sign change between -2 and -3, because f(-2)=7 and f(-3)=-1.

The one real root is between -2 and -3, therefore, it can not be an integer.

Actually, it's x=-2.925851551....

Just for fun, here's the factorization:

$\displaystyle (x+2.925851551)(x-0.4629257757+1.222539948i)$$\displaystyle (x-0.4629257757-1.222539948i)$

- Jan 11th 2007, 03:42 PM #5

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10