Factoring somewhat complex polynomial

Question

Can this be factored?

$$m^2(a-1) + 2m(a-1) -1$$

I have not been able to find any roots that will work when multiplied out. Does anyone see any options?

Answer 1

$$m^2(a-1)+2m(a-1)-1=(a-1)(m^2+2m)-1=(a-1)((m+1)^2-1)-1=(a-1)(m+1)^2-a$$ The roots of this equation would be $$(a-1)(m+1)^2-a=0\Rightarrow (m+1)^2=\frac{a}{a-1}\Rightarrow m=-1\pm\sqrt{\frac{a}{a-1}}$$ provided $a\notin (0,1)$.

Answer 2

This cannot be factored into polynomials in both $a$ and $m$, but

$$ m^2(a-1) + 2m(a-1) - 1 = \left((m+1)\sqrt{a-1}+\sqrt{a}\right)\left((m+1)\sqrt{a-1}-\sqrt{a}\right) $$