Given the equation:$$3x^4-25x^3+50x^2-kx+12=0$$ has two roots whose product is $2$. It was given to find the value of k and all the roots of the equation.This is how I solved it:
Let the roots be $$\alpha , \beta , \gamma , \delta$$ Let $$\alpha \beta=2$$So,$$\gamma \delta=2$$ $$\implies(x^2-(\alpha +\beta)x+2)(x^2-(\gamma +\delta)x+2)=0$$ Then by expanding we get :$$x^4-(\alpha+\beta+\gamma+\delta)x^3+(\alpha \gamma +\beta \gamma + \alpha \delta + \beta \delta +4)x^2-2(\alpha+\beta+\gamma+\delta)x+4=0$$ Comparing coefficients, we get $k=50$
Now,$$\alpha\gamma+\beta\gamma+\alpha\delta+\beta\delta+4=50/3$$ $$\implies (\alpha+\beta)(\gamma+\delta)=38/3$$ Let $$\alpha+\beta=k_1 $$ $$ \gamma+\delta=k_2$$ $$\implies k_1+k_2=25/3$$ & $$k_1 k_2=38/3$$ Solving first we get $$k_1 ,k_2$$ Then we get the roots as $$6,1/3,1+i,1-i$$ Now the question that arises is that can we generally solve the quartic by factorization if only a single information is given about its roots and any one of the coefficients is unknown?Also will the way I solved this question be always applicable or are there certain restrictions?
P.S: An explanation with elementary algebraic concepts will be preferred.