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.

Two of the roots sum to 4.  All 4 roots sum to 8.  This means that the other two roots must also sum to 4.
If this polynomial had rational roots, they would have to be in the set $\{\pm1,\pm5\}$
By what we have above we might try $x=-1, x = 5$ alas these do not work.
We could factor the polynomial like so:
$(x^2 - 4x + A)(x^2 -4x + B)\\
A+B = 5\\
AB = 5$
And solve for $A,B$
Or we might try
$(x -2 + a)(x-2-a)(x-2 + b)(x-2-b)$
Then substituting 
$y = x-2\\x = y+2$
Into the original polynomial
$(y+2)^4 - 8(y+2)^3 + 21(y+2)^2 -20(y+2) + 5\\
y^4 - 3y^2+1\\
y^2 = \frac 32 \pm \sqrt 2\\
y = \pm \sqrt {\frac 32 \pm \sqrt 2}\\
x = 2\pm \sqrt {\frac 32 \pm \sqrt 2}$
