I was wondering whether this expression could be factored by any means
$x^2+4x-1$
Many THanks
I was wondering whether this expression could be factored by any means
$x^2+4x-1$
Many THanks
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).
Or you could use the quadratic formula to show that $ \displaystyle x^2+ 4x- 1= 0$ has roots $ \displaystyle \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}$ so that its factors are $ \displaystyle (x- 2- \sqrt{5})(x- 2+ \sqrt{5})$.
Or "complete the square": $ \displaystyle x^2+ 4x- 1= x^2+ 4x+ 4- 4- 1= (x- 2)^2- 5= (x- 2)^2- (\sqrt{5})^2$, a "difference of two squares" which can be factored as a "sum and difference", $ \displaystyle (x- 2+ \sqrt{5})(x- 2- \sqrt{5})$.
($ \displaystyle a^2- b^2= (a+ b)(a- b)$)
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})$