Stuck on a simple factoring problem

Question

The answer to this question is probably very obvious but I can't figure it out for some reason:

I simply want to factorise: $x^2+5x-2$

I solve $x^2+5x-2 = 0$

i find $x_1 = \dfrac{-5-\sqrt{33}}{2}$ and $x_2 = \dfrac{5-\sqrt{33}}{2}$

If I want to factorise, I simply do $(x-x_1)(x-x_2) = (x - \dfrac{-5-\sqrt{33}}{2})(x-\dfrac{5-\sqrt{33}}{2})$

But when I check with Wolframalpha, they get: $-\dfrac{1}{4}(-2x+\sqrt{33} - 5)(2x+\sqrt{33} + 5)$

What am I missing

Answer 1

$x_{1,2}=\frac{-5\pm\sqrt{33}}{2}$ $$x^2+5x-2=(x - \dfrac{-5-\sqrt{33}}{2})(x-\dfrac{-5+\sqrt{33}}{2})=-\dfrac{1}{4}(-2x+\sqrt{33} - 5)(2x+\sqrt{33} + 5)$$

Answer 2

Your answer is almost correct, however through the quadratic formula one of your subtraction signs should have been an addition one.

Your answer should have been:

$$x^2+5x-2=(x - \dfrac{-5-\sqrt{33}}{2})(x-\dfrac{-5+\sqrt{33}}{2})$$

Then combine fractions:

$$(\frac{2x+5+\sqrt{33}}{2})(\frac{2x+5-\sqrt{33}}{2})$$

Thus:

$$\frac14(2x+5+\sqrt{33})(2x+5-\sqrt{33})$$

Or as Wolfram Alpha put it:

$$-\frac14(2x+5+\sqrt{33})(-2x-5+\sqrt{33})$$