Conference rusure::math

    if maple cannot factor it, what's the point of trying .. :-)
    > s;
                                      4      2
                                     x  + 4 x  + 5
    > factor(s);
                                      4      2
                                     x  + 4 x  + 5
    
    i cant wait to find the answer...
    /naser
    

        I know that x^4 + 4 factors as
        
        	x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2
        		= (x^2 + 2x + 2)(x^2 - 2x + 2)
        
        Applying the same thing to x^4 + Bx^2 + C^2 gives
        
        	x^4 + Bx^2 + C^2 = x^4 + 2Cx^2 + C^2 - (2C-B)x^2
        		= (x^2 + C)^2 - (sqrt(2C-B) x)^2	if 2C-B >= 0
        		= (x^2 + sqrt(2C-B) x + C)(x^2 - sqrt(2C-B) x + C)
        			if 2C-B >= 0
        
        So with the problem in .0,
        
        	ax^4 + bx^2 + c = a(x^4 + (b/a)x^2 + (c/a))
        
        Let B = b/a and C^2 = c/a.  We are given a>0 and b^2-4ac<0
        so we know 4ac > b^2 >= 0 and since a>0, then c>=0 as
        well.  So c/a >= 0 and we can write C^2 = c/a, B = b/a. 
        Also, 4ac - b^2 > 0 so 4c/a - (b/a)^2 > 0 so
        4c/a > (b/a)^2 >= 0 so 2sqrt(c/a) > |b|/a >= b/a.  But
        this is 2C >= B and so 2C - B >= 0 and we can use the
        previous formula, which gives
        
        	ax^4 + bx^2 + c = a(x^4 + (b/a)x^2 + (c/a))
        = a(x^2 + sqrt(2sqrt(c/a) - b/a) x + sqrt(c/a))
           (x^2 - sqrt(2sqrt(c/a) - b/a) x + sqrt(c/a))
        
        You can verify by multiplying it out; you get
        
        	a(x^4 + 2sqrt(c/a)x^2 + c/a - (2sqrt(c/a) - b/a) x^2)
        	= a(x^4 + b/a x^2 + c/a)
        	= ax^4 + bx^2 + c
        
        Dan

   An even polynomial can always be written in the form F(z)*F(-z).

   Consider

	(z^2 + p*z + 1)*(z^2 - p*z + q) =
	z^4 + (2*q-p^2)*z^2 + q^2 =
	z^4 + P*z^2 + Q

    Since you stipulate that P^2 < 4*Q, q = +sqrt(Q) and p = +sqrt(2*q-P)
    with no negative arguments appearing under the radicals.

    It's worth remarking that the roots of your quartic have quadrantal
    symmetry.

    - Jim

>Let a, b, and c be real numbers with a>0 and b^2-4ac<0.
>Express the polynomial ax^4+bx^2+c explicitly as the
>product of two quadratic polynomials with real coefficients.

	In fact, a factorization into real quadratics exists exactly when
a <> 0.   Wlog, take a = 1.   Then there are three factorizations:

	(i) & (ii) F(x)F(-x)

	(x�+px+q)(x�-px+q) = x^4 + bx� + c

=>	q� = c
	p� = 2q - b

	For p & q to be real here, first we require that c >= 0   Then
assuming this, we need to pick the sign of q such that 2q - b >= 0.   This
is possible if either of the following hold:
	(1) b =< 0
or	(2) b� =< 4c (which implies c >= 0)

	Alternatively
	(iii) F(x�)G(x�)

	(x�-k)(x�-l) = x^4 + bx� + c

=>	k,l = [-b � _/(b�-4c)] / 2

	For k & l to be real here, all we need is b� >= 4c.

	So depending on whether b�-4c < or > 0, we can't avoid having some
factorization into real quadratics.