Contrary to what's being said in the comments, neither of those expressions is wrong because both yield $a^4 + b^4$ when expanded. It is, however, incorrect to call this a factorization:
When you factor an expression, say $F(x)$, you break it down into two or more factors, such as $F(x) = G(x)H(x)K(x)$. That is not what you did here. You broke it down into two factors plus a remainder of $-2ab^3$. That is not factoring.
To address the question more generally, since $\Bbb R$ (or $\Bbb C$, or $\Bbb Q$, whichever you're working with) is a field, then $\Bbb R[x]$ is a unique factorization domain (among other things, but those other things are irrelevant here). This means that every expression in $\Bbb R[x]$ has one and only one factorization (up to ordering of the factors, for example $x(x-1)$ and $(x-1)x$ are considered the same factorization of $x^2-x$). An expression may have two factorizations that look different but actually aren't.
For a simple example over $\Bbb C[x]$, we could say $x^4 - 1 = (x^2 + 1)(x^2 - 1)$ and we could say $x^4 - 1 = (x^2 - x + i(1-x))(x^2 + x + i(1+x))$. These look very different but they're really the same. How can we be sure they're the same? Break it into linear factors over $\Bbb C$:
$$ x^4 - 1 = (x-1)(x+1)(x+i)(x-i)$$
$(x-1)(x+1) = x^2-1$ and $(x+i)(x-i) = x^2+1$. That's how we can get the first factorization.
Alternatively, $(x-1)(x-i) = x^2 - x + i(1-x)$ and $(x+1)(x+i) = x^2 + x + i(1+x)$. That's how we can get the second factorization.