1. ## 4th Degree Polynomial

How to factor this function?

y = x^4 + x^2 + 2

2. ## Re: 4th Degree Polynomial

Originally Posted by joshuaa
How to factor this function?

y = x^4 + x^2 + 2
what makes you think it will factor?

3. ## Re: 4th Degree Polynomial

the solution of the integral [ 1/(x^4 + x^2 + 2) ] dx makes me think it will factor

4. ## Re: 4th Degree Polynomial

Originally Posted by joshuaa
the solution of the integral [ 1/(x^4 + x^2 + 2) ] dx makes me think it will factor
what is the solution to that integral?

here's what Wolfram spits out ...

http://www.wolframalpha.com/input/?i...Bx%5E2%2B2)+dx

5. ## Re: 4th Degree Polynomial

The Solution Here Has no Imaginary i

6. ## Re: 4th Degree Polynomial

Instead of "factoring", I would "complete the square". $x^4+ x^2+ 2= (x^2)^2+ (x^2)+ \frac{1}{4}- \frac{1/4}+ 2= (x^2+ 1/2)^2+ \frac{7}{4}$. $x^2+ 1/2= x^2- (-1/2)= (x- i\sqrt{2}/2}(x+ i/sqrt{2}/2)$.

7. ## Re: 4th Degree Polynomial

Originally Posted by joshuaa

The Solution Here Has no Imaginary i
well, in either case it's not gonna be easy ...

8. ## Re: 4th Degree Polynomial

What are the steps to factor like the way below?

9. ## Re: 4th Degree Polynomial

$x^4+x^2+2 = (x^2+ax+b)(x^2+cx+d)$

$x^4+x^2+2 = x^4 +(a+c)x^3 +(b+ac+d)x^2 + (bc+ad)x + bd$

match coefficients ...

$a+c = 0$

$b+ac+d = 1$

$bc+ad = 0$

$bd = 2$

solve the system ... you can do that.

10. ## Re: 4th Degree Polynomial

c = -a
then
b - a^2 + d = 1 (equation 1)
-----
-ba = -ad mean b = d
------
back to equation 1
a = square root of (2b - 1)

NOW WHAT?

11. ## Re: 4th Degree Polynomial

$a+c=0 \implies a = -c$

$bc+ad=0 \implies bc - cd = 0 \implies c(b-d) = 0 \implies b = d$

$bd = 2 \implies b = d = \pm \sqrt{2}$

$b + ac+d = 1 \implies ac = 1-(b+d) \implies -c^2 = 1-(b+d) \implies c^2 = (b+d)-1 = 2b-1$ ... $c^2 > 0 \implies b=d=\sqrt{2}$ and $c = \pm \sqrt{2\sqrt{2}-1} \implies a=\mp \sqrt{2\sqrt{2}-1}$

$x^4+x^2+2 = (x^2 -\sqrt{2\sqrt{2}-1} x + \sqrt{2})(x^2 +\sqrt{2\sqrt{2}-1} x + \sqrt{2})$

12. ## Re: 4th Degree Polynomial

Thanks a lot.

Can I apply (x^2 + ax + b)(x^2 + cx + d) for any 4th degree polynomial?

13. ## Re: 4th Degree Polynomial

Originally Posted by joshuaa
Thanks a lot.

Can I apply (x^2 + ax + b)(x^2 + cx + d) for any 4th degree polynomial?
you can try ...

15. ## Re: 4th Degree Polynomial

any fourth degree polynomial with real coefficients and no real roots of the form

$x^4+A x^2+ B$

can be factored as follows ( $p,s$ real )

$\left(x^2-s x + p\right)\left(x^2+s\text{ }x +p\right)$

multply out to get

$x^4+\left(2p-s^2\right)x^2+p^2$

and compare coefficients and note that $B$ must be positive otherwise the polynomial has real roots