MHB Can this expression be factored?

[mathlearn]

I was wondering whether this expression could be factored by any means

$x^2+4x-1$Many THanks :)

 

[Ackbach]

Not over the rationals, but if you multiply out $\left(x+\sqrt{5}+2\right)\left(x-\sqrt{5}+2\right)$, I fancy you'll get your original expression back. Wolfram|Alpha is a great way to check these sorts of things. You can use the quadratic formula to get it by hand (set your quadratic equal to zero).

 

[HallsofIvy]

Or you could use the quadratic formula to show that $$x^2+ 4x- 1= 0$$ has roots [math]\frac{-4\pm\sqrt{4^2- 4(1)(-1)}}{2(1)}= \frac{-4\pm\sqrt{16+ 4}}{2}= \frac{-4\pm\sqrt{20}}{2}= \frac{4\pm2\sqrt{5}}{2}= 2\pm\sqrt{5}[/math] so that its factors are [math](x- 2- \sqrt{5})(x- 2+ \sqrt{5})[/math].

Or "complete the square": [math]x^2+ 4x- 1= x^2+ 4x+ 4- 4- 1= (x- 2)^2- 5= (x- 2)^2- (\sqrt{5})^2[/math], a "difference of two squares" which can be factored as a "sum and difference", [math](x- 2+ \sqrt{5})(x- 2- \sqrt{5})[/math].
([math]a^2- b^2= (a+ b)(a- b)[/math])

 

[bwpbruce]

Complete The Square Method:
$x^2 + 4x - 1 $
$= x^2 + 4x + 4 - 4 - 1$
$ = (x + 2)^2 - 5$
$ = (x + 2)^2 - (\sqrt{5})^2$
$ = (x + 2 + \sqrt{5})(x + 2 - \sqrt{5})$