Factoring x^4 - 14x^2 + 52 over the reals

[menager31]

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

 

[morphism]

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

 

[menager31]

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

 

[menager31]

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

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

 

[robert Ihnot]

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

 

[D H]

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

x^4-14x^2+52

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

 

[CRGreathouse]

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

Isn't that statement equivalent to the Mertens conjecture?

 

[Hurkyl]

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.​

 

[genneth]

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

 

[D H]

As the equation has no real roots, you are looking for the product of a pair of quadratics.

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

You don't need a and d. Since ad=1, you can scale the two polynomials to make a and d equal to 1.

ae+bd=1

This is the source of your problems. Try again.

 

[D H]

Therefore, ...

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

 

[genneth]

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

Apologies -- got lazy.