$let\;\lambda,\mu\in\mathbb{Q}$ 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 $$ \gcd\left (2x^3-7x^2+2x+3,x^3-3x^2-x+3\right)=x^2-4x+3. $$

  • $\begingroup$ Look at the edit again. The answer cannot that it works for any rational $\lambda and \mu.$ $\endgroup$ – C. Cristi Apr 7 '18 at 8:33
  • $\begingroup$ Yes, you are right. I had a typo. $\endgroup$ – Dietrich Burde Apr 7 '18 at 11:35
  • $\begingroup$ How did you got there? $\endgroup$ – C. Cristi Apr 7 '18 at 14:46
  • $\begingroup$ For the other values in a)-f) the gcd is obviously equal to $1$. I just divided one polynomial by the other. $\endgroup$ – Dietrich Burde Apr 7 '18 at 14:50
  • $\begingroup$ so you tried all the answers? that's a bit of a unorthodox $\endgroup$ – C. Cristi Apr 7 '18 at 14:55

Hint: $$2x^3-7x^2+ \lambda x+3 = 2(x- \alpha)(x - \beta)(x-\gamma)$$
$$x^3 - 3x^2 + \mu x + 3 = (x- \alpha)(x - \beta)(x-\delta)$$

Now use vieta's formula to find $\alpha $ and $\beta$.

  • $\begingroup$ @DietrichBurde if G.C.D. is a polynomial of order 2 then it must be divisible by that polynomial which have root $\alpha$ and $\beta$ $\endgroup$ – kayush Apr 7 '18 at 8:15
  • $\begingroup$ @DietrichBurde i am not sure but rank 2 $\neq$ degree2? $\endgroup$ – kayush Apr 7 '18 at 8:20
  • $\begingroup$ @KumarAyush rank 2 = degree $\endgroup$ – C. Cristi Apr 7 '18 at 8:20
  • $\begingroup$ @C.Cristi didn't understand ur comment $\endgroup$ – kayush Apr 7 '18 at 8:22
  • $\begingroup$ @DietrichBurde I edited the question please look. $\endgroup$ – C. Cristi Apr 7 '18 at 8:26

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