Find coefficient in quartic given product of roots The product of two of the roots of 
$$x^4 -11x^3 + kx^2 + 269x - 2001=0$$  is $-69$.
Find k.
This is a question I have recently received, and I am required to take a test on related questions tomorrow. I have absolutely no idea on how to start with this or even which concept this belongs to. I would greatly appreciate it if someone tells me a link online to learn the theory required to answer this question and a page where I can learn how to solve questions similar to this. A solution would also be appreciated. I am only in 9th grade so I am absolutely clueless.
Apologies for the amateurish question.
 A: Suppose the roots are $a,b,c$ and $d.$ Then by  Vietas formulas, $$a+b+c+d=11\to(1)$$ $$ab+ac+ad+bc+bd+cd=k\to(2)$$ $$abc+abd+acd+bcd=-269\to(3)$$ $$abcd=-2001\to(4).$$ By your additional condition (suppose) $$ab=-69.$$ Then by $(4)$$$cd=29$$ and by $(3)$ $$ab(c+d)+cd(a+b)=-69(c+d)+29(a+b)=-269.$$ $$-69(c+d)+29(11-(c+d))=-269.$$ Hence $$c+d=6$$ and $$a+b=5.$$ Finally $$ab+ac+ad+bc+bd+cd=ab+(a+b)(c+d)+cd=k.$$
Hence $k=-10.$
A: The following is basically the answer by Nilan, reformulated to avoid explicit references to each of the $4$ roots (since in any case one cannot distinguish the two whose product is given form each other, nor the remaining two from each other).
Your polynomial must factor as product of two monic quadratic polynomials, the first one with the two roots whose product is $-69$. This gives an equation
$$
  x^4 -11x^3 + kx^2 + 269x - 2001 = (x^2 + px -69)(x^2+qx+r)
$$
where $p,q,r$ are to be determined. Comparing coefficients gives the equations
$$
  \begin{align}
  -11 &= p+q, \\
   k  &= r+pq-69, \\
 269  &= pr - 69 q, \\
-2001 &= -69r.
  \end{align}
$$
The final equation gives $r=29$. Then the first and third equations become a linear system in $p,q$, whose solution is $p=-5,q=-6$. Finally the second equation gives $k=29+30-69=-10$.
