# The hardest polynomial

1. Sep 27, 2007

### menager31

x^4-14x^2+52
i don't know how to factorize it in reals.

2. Sep 27, 2007

### morphism

Maybe because it can't be factored over the reals?

3. Sep 27, 2007

### menager31

but i read that all pol. can be factored in reals and the higher power of x can be 2

4. Sep 27, 2007

### menager31

(ax2+bx+c)(dx2+ex+f)

ae+bd=1
cf=52
bf+ce=0
be+af+dc=-14
c,a,d,f =/= 0

5. Sep 27, 2007

### robert Ihnot

X^2 +1=0, this polynominal can be factored over the reals?

6. Sep 27, 2007

### Staff: Mentor

All polynomials can be factored in the complex numbers, not the reals.

This has no real roots. Hint: The expression is quadratic in x2.

7. Sep 27, 2007

### CRGreathouse

Isn't that statement equivalent to the Mertens conjecture?

8. Sep 27, 2007

### Hurkyl

Staff Emeritus
Fundamental theorem of real algebra:
Every monic polynomial can be uniquely factored into a product of monic irreducible polynomials. Any irreducible polynomial is either linear or quadratic.​

Last edited: Sep 27, 2007
9. Sep 27, 2007

### genneth

Guys, he's saying that all polynomials with real coefficients can be factors as (at most) quadratics with real coefficients. This is true.

Last edited by a moderator: Sep 27, 2007
10. Sep 27, 2007

### Staff: Mentor

As the equation has no real roots, you are looking for the product of a pair of quadratics.
You don't need a and d. Since ad=1, you can scale the two polynomials to make a and d equal to 1.

This is the source of your problems. Try again.

11. Sep 27, 2007

### Staff: Mentor

Did you read the guidelines? Don't post complete solutions.

12. Sep 27, 2007

### genneth

Apologies -- got lazy.