Factorisation of $\frac{1}{u^2+3u-1}$

Question

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?

Answer 1

$$\frac{1}{(x-x_1)(x-x_2)} = \frac{1}{(x - \frac{-3+\sqrt{13}}{2})(x-\frac{3+\sqrt{13}}{2})} = \frac{1}{(\frac{2x + 3-\sqrt{13}}{2})(\frac{2x -3-\sqrt{13}}{2})}=\frac{4}{(2x + 3-\sqrt{13})(2x - 3-\sqrt{13})}$$

Answer 2

If you look carefully, multiplying out the factors in your denominator will lead to an extra factor of $4$, which is easily noticeable on the quadratic term...

This factor appeared from you substituting $$ (x - x_1)(x-x_2) \leftrightarrow (2x-2x_1)(2x-2x_2) $$ in your factorization.

Answer 3

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