If I have a polynomial $x^2(1-m^2) - x2m^2 - (m^2 + 1)$ with a solution at $x = -1$, how do I get the other root

Question

If I have a polynomial $x^2(1-m^2) - x2m^2 - (m^2 + 1)$ with a solution at $x = -1$, then I know I can just take $x^2(1-m^2) - x2m^2 - (m^2 + 1)$ and divide it by $x+1$ to get the other root.

In a similar manner, if I have $x^2(m^2 + 1) + x(2m-2m^2) + (m^2 - 1)$ with a solution at $x = 1$, then I know I can just take $x^2(m^2 + 1) + x(2m-2m^2) + (m^2 - 1)$ and divide it by $x-1$ to get the other root.

Problem is when I tried long division here for both, I get such a ridiculous mess that I don't know how to move on. In a similar fashion, wolframalpha doesn't help either. Just in case I my polynomials are wrong, here is where I derived them from:

For the first one I had the two equations: $y = m(x-1) + 1$ and $x^2 + y^2 = 2$ so I put the first one into the second one and simplified.

For the second one I had the two equations: $y = m(x+1)$ and $x^2 - y^2 = 1$. Similarly I put the first one into the second one and simplified.

I would be very grateful if I could get the second root of each one.

Answer 1

For the quadratic equation $(1-m^2)x^2-2m^2 x-(m^2+1)=0$, the product of the roots is $\frac{-(m^2+1)}{1-m^2}$. Since $-1$ is one of the roots, the other must be $\frac{1+m^2}{1-m^2}$.

The same method can be used for any quadratic equation with one known root.

Remark: Please check the calculation of the coefficients of the quadratics. For example, when we substitute $y=m(x-1)+1$ in $x^2+y^2=2$, we do not get your first quadratic, nor your second, though that one is much closer.

The equations $y=m(x-1)+1$ and $x^2+y^2=2$ were written down in order to give a rational parametrization of the quadratic. If you want to use $y=m(x-1)+1$, then one of the roots of the resulting quadratic will be $1$, not $-1$.

Answer 2

To do polynomial division in such cases, you may want to consider the following approach: I shall illustrate with your title problem.

-1  | (1-m^2)    -2m^2  -(m^2+1)
    |    0       -1+m^2    1+m^2               
    -------------------------------
       1-m^2     -1-m^2   |  0

This gives you the factorisation $(x+1)\left((1-m^2)x-(1+m^2)\right)$ from which the second root is clear.