<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] You know that $abcd = -8$ and $cd = -8.$ Therefore $ab= 1.$
From the coefficient of $x$ in the polynomial, you know that $abc + acd + abd + bcd = -26$ (note: not $+26$).
But since $ab=1$ and $cd=-8$, you can see that
$abc + acd + abd + bcd = c - 8a + d - 8b.$
Therefore
$$ -8a - 8b + c + d = -26. \tag1 $$
But you also know that
$$ a + b + c + d = 1. \tag2 $$
Subtract Equation $(1)$ from Equation $(2)$:
$$ 9a + 9b = 27. $$
That is, $a + b = 3.$  But $b = \frac1a,$ so we have
\begin{align}
a + \frac1a &= 3, \\
a^2 + 1 &= 3a, \\
a^2 - 3a + 1 &= 0. \\
\end{align}
Apply the quadratic formula to solve $y^2 - 3y + 1 = 0.$
Note by symmetry that $a$ and $b$ both are solutions of this equation.
But you are given that $a > b$, so you can see how to match $a$ and $b$ with the two solutions of the quadratic formula.
For $c$ and $d,$ multiply Equation $(2)$ by $8$ and add the result to Equation $(1)$.
You get $9c + 9d = -18.$ But also $d = -\frac8c.$ Again you can get a quadratic equation out of this and solve it, then use the information that $c < d$ to know which root is which.
No guesswork is required, although if you do guess cleverly you can shorten the path a little.
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