How to factor this function?
y = x^4 + x^2 + 2
what is the solution to that integral?
here's what Wolfram spits out ...
http://www.wolframalpha.com/input/?i...Bx%5E2%2B2)+dx
Instead of "factoring", I would "complete the square". $\displaystyle x^4+ x^2+ 2= (x^2)^2+ (x^2)+ \frac{1}{4}- \frac{1/4}+ 2= (x^2+ 1/2)^2+ \frac{7}{4}$. $\displaystyle x^2+ 1/2= x^2- (-1/2)= (x- i\sqrt{2}/2}(x+ i/sqrt{2}/2)$.
$a+c=0 \implies a = -c$
$bc+ad=0 \implies bc - cd = 0 \implies c(b-d) = 0 \implies b = d$
$bd = 2 \implies b = d = \pm \sqrt{2}$
$b + ac+d = 1 \implies ac = 1-(b+d) \implies -c^2 = 1-(b+d) \implies c^2 = (b+d)-1 = 2b-1$ ... $c^2 > 0 \implies b=d=\sqrt{2}$ and $c = \pm \sqrt{2\sqrt{2}-1} \implies a=\mp \sqrt{2\sqrt{2}-1}$
$x^4+x^2+2 = (x^2 -\sqrt{2\sqrt{2}-1} x + \sqrt{2})(x^2 +\sqrt{2\sqrt{2}-1} x + \sqrt{2})$
any fourth degree polynomial with real coefficients and no real roots of the form
$\displaystyle x^4+A x^2+ B$
can be factored as follows ($\displaystyle p,s$ real )
$\displaystyle \left(x^2-s x + p\right)\left(x^2+s\text{ }x +p\right)$
multply out to get
$\displaystyle x^4+\left(2p-s^2\right)x^2+p^2$
and compare coefficients and note that $\displaystyle B$ must be positive otherwise the polynomial has real roots