<s>[INST] Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrightarrow 4ab + 4cd = 20 \Longrightarrow ab + cd = 5$$
$$abcd = 5$$
I can't figure out how to proceed.
 [/INST] $x^4-8x^3+21x^2-20x+5=(x^2-4x+a)(x^2-4x+b)$
$\begin{cases}a+b+16=21 \\ -4a-4b=-20\\
ab=5 \end{cases}$
So $a+b=5$ and $ab=5$.
$a$, $b$ are the roots of $t^2-5t+5=0$.
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