<s>[INST] How to factorize polynomials to the 5th degree? I have the polynomial:
$$2x^5-x^4+10x^3-5x^2+8x-4$$
and I know that the final result is:
$$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$
But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a common factor between the expressions but in this case their are none.
 [/INST] Laboring under the assumption that such a polynomial is not irreducible and can be factored (which is clear in this case since you already have a factorization), we would generally start by looking for rational roots.
The rational root theorem asserts that if $z = \frac{a}{b}$ is a rational root of the polynomial (in lowest terms), then $b$ must divide the leading coefficient, and $a$ must divide the constant term.  This gives us several potential rational roots:
$$ \left\{ \pm 1, \pm 2, \pm 4, \pm \frac{1}{2} \right\}. $$
By trial and error, it is possible to determine that the only one of these that is, in fact, a root is $\frac{1}{2}$.  From this, we know that
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right) q(x), $$
where $q$ is a polynomial of degree 4.  Via polynomial long division, Horner's algorithm / synthetic division, or some other technique, we can deduce that $q(x) = 2x^4 + 10x^2 + 8$, thus
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4 = \left( x - \frac{1}{2} \right)\left( 2x^4 + 10x^2 + 8 \right). $$
The last term is quadratic in $x^2$.  Via the quadratic formula, completing the square, or some other technique, we can determine that it, too, factors.  For example, we can "factor by grouping:"
$$ 2x^4 + 10x^2 + 8
= 2x^4 + 8x^2 + 2x^2 + 8
= 2x^2(x^2 + 4) + 2(x^2+4)
= (2x^2 + 2)(x^2 + 4).
$$
Note that both of these factors are irreducible over $\mathbb{R}$ (they have complex roots, but no purely real roots), hence we can stop here.  Putting all of the pieces back together, we get
$$ 2x^5 - x^4 + 10x^3 - 5x^2 + 8x - 4
= \left( x - \frac{1}{2} \right) (2x^2 + 2)(x^2 + 4),
$$
which is equivalent to the factorization you gave.
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