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.
Every now and then, you find a polynomial of higher degree that can be factored by inspection. In this case, there's a way to just "see" one step of the factorization:

$$2x^5-x^4+10x^3-5x^2+8x-4$$

Notice that the coefficients, when grouped in pairs, are all proportional: $2, -1$ are in the same ratio as $10,-5$ and also $8,-4$. That's uncommon, but it means you can "factor by grouping":

$$(2x^5-x^4)+(10x^3-5x^2)+(8x-4)$$
$$=x^4(2x-1)+5x^2(2x-1)+4(2x-1)$$
$$=(x^4+5x^2+4)(2x-1)$$

The first step is to take the gcf out of each group, and then, since the remaining parts match ($(2x-1)$), that can be factored out. Now we're left with something that is quadratic in form, and can be factored further:

$$x^4+5x^2+4=(x^2+4)(x^2+1)$$



Now, in most cases, nice patterns like this do not occur. Then, you're left with the rational roots theorem. From the fact that our lead coefficient is $2$ and our constant term is $4$, we conclude that any linear factor will correspond to a zero of the form $\pm\frac{p}{q}$, where $p$ is a factor of $4$ and $q$ is a factor of $2$. That gives us a few options: $\pm 4, \pm 2, \pm 1, \pm \frac12$. Trying a few of these, we see that plugging in $x=\frac12$ actually produces a value of $0$. That tells us that $(x-\frac12)$ is a factor, or clearing denominators, $(2x-1)$. You can now factor this linear piece out by long division, and end up just where we were after factoring by grouping.
The question slightly wrong in my opinion. Please try multiplying the factors or the eqn. and lets see if you will get back the original question.

The correct equation is $P(x): 2x^5-x^4-10x^3+5x^2+8x-4$

Group and Factorize: $(2x^5-x^4) (-10x^3+5x^2) (8x-4) =0$

                 $x^4(2x-1)   -5x^2(2x-1)  4(2x-1)$

        Factor out $(2x-1)$ then we have $x^4-5x^2+4$


Our first factor of the equation is $(2x-1)$

This looks like a quadratic form but our first coefficient has 4th degree

look let's expand $x^4-5x^2+4$ and we have $(x^2-4)$ and $(x^2-1)$. $-4$ and $-1$ when added will be $-4-1 = -5$ and when multiply $-4x-1 = 4. x^2 x x^2 = x^4$
so its just splitting them and distributing them on $-4$ and $-1$ which are the product of the constant and sum of the middle term.

Now with $(x^2-4)$ and $(x^2-1)$ we can have $(x-2) (x+2)$ and $(x-1) (x+1)$.

so our final factors of the polynomial are $(2x-1) (x-2) (x+2) (x-1) (x+1)$

Therefore; the zeros of the polynomial are $1/2, 2, -2, 1, -1$.