$let\;\lambda,\mu\in\mathbb{Q}$ such that the GCD($2X^3-7X^2+\lambda X+3, X^3-3X^2+ \mu X +3)$ IS A POLYNOMIAL OF DEGREE 2.
I know this problem may seem trivial to you, but for me it seems like I just can't get it right with the greatest common divisor. I tried to use euclid's algorithm.
First I divided $\frac {2X^3-7X^2+\lambda X+3}{X^3-3X^2+ \mu X +3}$ and obtained a remainder of: $-X^2+(\lambda -2\mu)X-3$ then continued with the algorithm... I divided $\frac {X^3-3X^2+ \mu X +3}{-X^2+(\lambda -2\mu)X-3}$ and obtained a remainder of $(\lambda-2\mu-3)X^2+(\mu-3)X+3$, then using the problem's condition i thought this should be equal to 0 but it can't be.. because of the constant.. so then I divided again $\frac{-X^2+(\lambda -2\mu)X-3}{(\lambda-2\mu-3)X^2+(\mu-3)X+3}$ and now the remainder i obtained was more pleasant to the problem's condition. It was: $(\frac{\mu-3}{\lambda-2\mu-3}+\lambda -2\mu)X+ \frac {3}{\lambda-2\mu-3}-3$ and then I set X coefficient to equal $0$ and the constant term to equal to $0$ too such that it's GCD would be a polynomial of degree 2.
Got into a 2x2 system but the solutions were no good.
Where do I go wrong?
In my book I have answer posibilities so the posibilities are:
a) $\lambda = -1, \mu = 2$ b) $\lambda = 0, \mu = 0$ c) $\lambda = 2, \mu = 0$
d) $\lambda = 2, \mu = -1$ e) $\lambda = -1, \mu = -1$ f) $\lambda = 0, \mu = 2$
Option $d$ is correct: we have $$ \gcd\left (2x^3-7x^2+2x+3,x^3-3x^2-x+3\right)=x^2-4x+3. $$