Can You Factor Me?

We are given the following biquadratic equation

$x^4 - 18x^3 + kx^2 + 174x - 2015 = 0$

Let the four roots of the equation be $\alpha, \beta, \gamma, \delta$

According to the question,

$\alpha\beta=-31$

On comparing the coefficients of the given equation with the standard form of the biquadratic equation(i.e. $ax^4 + bx^3 + cx^2 + dx - e = 0$ ), we find that,

$a=1,\quad b=-18,\quad c=k,\quad d=174,\quad e=-2015$

As per theory of equations we get,

product of all roots $= \alpha \beta \gamma \delta =\frac{e}{a}=-2015$

$\Rightarrow (-31)\gamma \delta = -2015$

$\therefore \quad \gamma \delta =65$

We also know,

$\sum { \alpha \beta \gamma } =174$

$\Rightarrow \quad \alpha \beta \gamma +\beta \gamma \delta +\alpha \gamma \delta +\alpha \beta \delta =174\\ \Rightarrow \quad -31\gamma +65\beta +65\alpha -31\delta =174\\ \Rightarrow \quad 65(\alpha +\beta )-31(\gamma +\delta )=174\quad \quad ...(1)$

Also,

$\sum { \alpha } =\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =-(-18)\\ \Rightarrow \alpha +\beta =18-(\gamma +\delta )$

Now, putting $\alpha +\beta =18-(\gamma +\delta )$ in $eq (1)$

$\\ 65(18-(\gamma +\delta ))-31(\gamma +\delta )=174\\ \Rightarrow \quad 1170-96(\gamma +\delta )=174\\ \Rightarrow \quad -96(\gamma +\delta )=-1344\\ \Rightarrow \quad \gamma +\delta =14$

$\therefore \quad (\alpha +\beta )=18-(\gamma +\delta )=18-14=4$

Now,

$\sum { \alpha \beta } =\alpha \beta +\beta \gamma +\gamma \delta +\alpha \delta +\delta \beta +\alpha \gamma =\frac { c }{ a } =k\\ \Rightarrow -31+\beta \gamma +65+\alpha \delta +\delta \beta +\alpha \gamma \\ =34+\alpha (\delta +\gamma )+\beta (\gamma +\delta )=k\\ \Rightarrow 34+(\alpha +\beta )(\gamma +\delta )=k\\ \Rightarrow 34+4(14)=k\\ \therefore k=\boxed{90}$