Solve the biquadratic!

METHOD 1

Let $f(x) = x^{4}-20x^{3}+kx^{2}+590x-1992$ and let $a,b,c,d$ are the roots of the given equation. Then by Vieta's Formula

$a + b + c +d = 20$ and $abcd = -1992$

WLOG let $ab = 24$ , then $cd = -83$ . Also by Vieta's Formula we have

$abc + abd + acd + bcd = -590$

$ab(c +d) + cd (a +b ) = - 590$

$24( 20 - (a + b)) - 83 (a + b) = - 590$

$480 - 107 (a + b) = - 590$

$\implies a + b = 10$

With $ab = 24$ and $a + b = 10$ , we get $a = 6$ and $b = 4$ or vice versa

Now by factor theorem $f(4) = 0 \implies 4^{4}-20\times 4^{3}+k\times 4^{2}+590\times 4 -1992 = 0 \implies k = \boxed{41}$

METHOD 2

The given equation is equal to $(x^2 + a x + 24)(x^2 + b x - 83)$ . Equating coefficients of $x^3$ and $x$ gives two equations in a and b, that is $a + b = -20$ and $-83 a + 24 b = 590$ .

Solving them gives $a = b = -10$ . Putting in the above equation to find coefficient of $x^2$ which is $k = \boxed{41}$