Can p^4 + 4 be factored? What about x^2 + 1?

[mathdad]

Factor p^4 + 4.

I know that this cannot be factored but don't know the reason the expression cannot be factored. Can someone explain why it cannot be factored?

 

[MarkFL]

If we allow for complex factors, then we can state:

\(\displaystyle p^4+4=\left(p^2+2i\right)\left(p^2-2i\right)\)

However, over the reals, the sum of two squares is not factorable. To see why, consider:

\(\displaystyle a^2+b^2=0\) where $a$ and $b$ are non-zero real numbers.

If we subtract through by $b^2$, we have:

\(\displaystyle a^2=-b^2\)

The square of a real number can never be negative, so there is no factorization of the sum of two real squares over the reals and thus there are no real solutions here...what we find instead is:

\(\displaystyle a=\pm bi\)

 

[kaliprasad]

Factor p^4 + 4.

I know that this cannot be factored but don't know the reason the expression cannot be factored. Can someone explain why it cannot be factored?

No it can be factored

$p^4+4 = p^4 + 4p^2 + 4 - 4p^2 = (p^2+2)^2 - (2p)^2 = (p^2 + 2p +2)(p^2 -2 p +2)$

 

[mathdad]

Can we say that p^4 + 4 is irreducible?

 

[Euge]

Can we say that p^4 + 4 is irreducible?

No. It is reducible, since $p^2 - 2p + 2$ and $p^2 + 2p + 2$ are factors of $p^4 + 4$, as kaliprasad showed.

More generally, a polynomial with real coefficients is called reducible if it can be written as a product of two real polynomials of lesser degree.

 

[mathdad]

How about x^2 + 1? Is this irreducible? If so, why?