<s>[INST] Factorisation of $\frac{1}{u^2+3u-1}$ So, I need to factor the expression $\frac{1}{u^2+3u-1}$ First I find the roots $x_1=\frac{-3+\sqrt{13}}{2}$ and $x_2=\frac{-3-\sqrt{13}}{2}$ then I have $\frac{1}{(2x+3+\sqrt{13})(2x+3-\sqrt{13})}$ But on the factorisation calculator it states that there's 4 in numerator. So my question is why?
 [/INST] $$\frac{1}{(x-x_1)(x-x_2)} = \frac{1}{(x - \frac{-3+\sqrt{13}}{2})(x-\frac{3+\sqrt{13}}{2})} = \frac{1}{(\frac{2x + 3-\sqrt{13}}{2})(\frac{2x -3-\sqrt{13}}{2})}=\frac{4}{(2x + 3-\sqrt{13})(2x - 3-\sqrt{13})}$$
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