I was trying to solve the problem:
Given that a and b are distinct positive numbers, find a polynomial P(x) such that the derivatives of $P(x)e^{-x^2}$ is zero for $x=0, x=a, x=-a, x=b, x=-b$.
We need to find the polynomial of p(x) in terms of a and b obviously.
What I did is:
We know $P’(x) - 2xP(x)=kx(x^2-a^2)(x^2-b^2)$.
Now, the next step is to assume the terms in P(X). So what I did is I assume:
$P(x)=\alpha x^4 + \beta x^3 + cx^2 + dx + g$
However, I saw the mark scheme, the P(x) should be assumed to be:
$x^4 +\beta x^3 +cx^2 + dx +g$, which means the coefficient should be 1 in front of the $x^4$. May I know why is this so? Or it is still be corrected if I put a constant in front of the $x^4$ term.
Thank you very much for your reply.
If you find a $P(x)$ that works, any constant times $P(x)$ will work as well because it will just multiply all the $0$s you are producing. You might as well divide by the coefficient of $x^4$ for standardization. Alternately, you could delete $k$ from your expression, which would give perhaps a different standardization.
You have six parameters in the solution, $\alpha, \beta, c,d,g,k$ and five equations for the degrees $0$ through $4$ in your equation, so if you do not delete $\alpha$ or $k$ (or one of the others) you will not have a unique solution.