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?

$$\frac{1}{{{u}^{2}}+3u-1}=\frac{1}{{{u}^{2}}+3u+\frac{9}{4}-\frac{13}{4}}=\frac{1}{{{\left( u+\frac{3}{2} \right)}^{2}}-\frac{13}{4}}=\frac{4}{{{\left( 2u+3 \right)}^{2}}-13}$$
