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?
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asked Oct 19, 2014 at 22:35
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$\begingroup$ Is this a polynomial in $m$ or in 2 variables? $\endgroup$– N. S.Oct 19, 2014 at 22:45
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2 Answers
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$$m^2(a-1)+2m(a-1)-1=(a-1)(m^2+2m)-1=(a-1)((m+1)^2-1)-1=(a-1)(m+1)^2-a$$ 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 $a\notin (0,1)$.
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$\begingroup$ Shouldn't it be $m = -1\pm\sqrt{\frac{a}{a-1}}$? $\endgroup$– EffOct 19, 2014 at 22:58
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This cannot be factored into polynomials in both $a$ and $m$, but
$$ m^2(a-1) + 2m(a-1) - 1 = \left((m+1)\sqrt{a-1}+\sqrt{a}\right)\left((m+1)\sqrt{a-1}-\sqrt{a}\right) $$
answered Oct 19, 2014 at 22:53