Factorizing $x^4-13x^3+56x^2-19$ I have been trying to factor the following quartic equation for two days now and I cannot find any way to factor it:

$$x^4-13x^3+56x^2-19.$$

I just can't seem to factor it at all. Is this equation able to be factored at all? Thanks in advanced guys!
.By factoring the quartic, I assume you want me to express the given polynomial as a product of two or more irreducible integer coefficient polynomials. I shall do more : I shall try to provide the roots of this polynomial as well.

We claim that this polynomial cannot be written as a product of integer polynomials of strictly smaller degree! We make use of a nice criteria, which you may ignore if you do not know : a polynomial with integer polynomials is irreducible if there is a prime $p$ such that when we reduce the coefficients of our polynomial modulo $p$, the new polynomial is irreducible in the finite field $\mathbb F_p$. 

If we do this with $p=2$, we get the polynomial $x^4+x^3+1$. This is much easier to check for irreducibility ,actually : you can check it has no linear factors by substituting $0$ and $1$ in, and the only irreducible polynomial in degree $2$ is $x^2+x+1$, which can be seen by the division algorithm to not divide this.

Hence, if you are seeking to factorize your polynomial as a product of integer polynomials, you cannot do this! 

But, you can still go in for "root finding" i.e. locating suitable regions where the (complex) roots of this polynomial are contained.

To do this, we make use first, of the Descartes' rule of signs. You may read this up in the link below[1].

Using the rule of signs of coefficients, there are three sign changes, and therefore either three or one positive root(s). Furthermore, there is exactly one negative root, by applying the rule to the negative of the above polynomial.

However,  there is an improvement of the Descartes' rule : instead of taking the sign changes of coefficients, you take the partial sums in the reverse order of coefficients : i.e. here you'd get $-19,-19,37,24,25$. There's precisely one change of sign, and therefore exactly one positive root.

Now, there's exactly one positive and one negative root, so where are these? Clearly, $f(-1) &gt;0 ,f(0)&lt;0,f(1)&gt;0$ by substitution, so by the intermediate value theorem we get that both have modulus less than $1$ and one is positive, the other negative.

Finally, as to the complex roots, we have the wonderful Rouche's theorem, which (I'll be brief) uses the fact that $56$ is greater than the sum of absolute values of all other coefficients, to conclude that two roots of $f$ lie inside the complex unit circle, and  two outside. Therefore both complex roots lie outside the unit circle.

I believe that there should be easily derivable upper bounds for the moduli of the complex roots as well, but at the moment this is all I am able to say, so I'd be happy if somebody can make this tighter.

[1] : 
 https://pdfs.semanticscholar.org/bc6c/59cbf7e5f76719e85afa669ce238ab56b80d.pdf
You find all possible roots of your expression.

To do this, you can use the rational root theorem and plug values of the possible rational roots in, which are $\pm19, \pm1$.

Another way you can do this is with graphing, most of the time, a graph with decimals to the third place is approximated and cannot be expressed as a fraction.

Therefore, there is no way to factor this, as its factors cannot be expressed in the form $x-\cfrac pq$,

where $\cfrac pq$ is some rational number.

Another quartic term which cannot be factored is $x^4-3$, but you know the roots from looking at the expression. Such is not the case in your expression.