<s>[INST] Factorize polynomial of degree 4 given statements If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?

Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$
Can I solve $a,b,c,d$ from there? How can I do so?
 [/INST] Let $$f(x) = x^4-x^3-13x^2+26x-8$$
Observe that $f(2) = 0$ and $f(-4) = 0$
Hence $(x-2)$ and $(x+4)$ are two factors of $f(x)$
Also $f(x)$ is divisible by $(x-2)(x+4) = x^2 + 2x - 8$
Now divide $f(x)$ by $x^2 + 2x - 8$ to obtain the other quadratic factor $x^2 - 3x +1$
If you solve $x^2 - 3x +1 = 0$, you'll find $x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$
Finally
$$f(x) = \left(x - \left(\frac{3}{2} + \frac{\sqrt{5}}{2}\right) \right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) \right) (x + 4)(x-2)$$
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