Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$

Question

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.

Answer 1

$x^4-8x^3+21x^2-20x+5=(x^2-4x+a)(x^2-4x+b)$

$\begin{cases}a+b+16=21 \\ -4a-4b=-20\\ ab=5 \end{cases}$

So $a+b=5$ and $ab=5$.

$a$, $b$ are the roots of $t^2-5t+5=0$.

Answer 2

make the Ansatz $$x^4-8x^3+21x^2-20x+5=(x^2+Ax+B)(x^2+Cx+D)$$

Answer 3

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

$(y+2)^4 - 8(y+2)^3 + 21(y+2)^2 -20(y+2) + 5\\ y^4 - 3y^2+1\\ y^2 = \frac 32 \pm \sqrt 2\\ y = \pm \sqrt {\frac 32 \pm \sqrt 2}\\ x = 2\pm \sqrt {\frac 32 \pm \sqrt 2}$