Factoring a 3-Variable Polynomial
Date: 08/13/99 at 06:01:33 From: Daniel Baumgartner Subject: Factorization in complex field Is it possible to factor the polynomial k1*x^2 + k2*y^2 + k3*z^2 + k4*xy + k5*xz + k6*yz into (g1*x + g2*y + g3*z)(h1*x + h2*y + h3*z) over the complex field for every choice of k1, ..., k6?
Date: 08/13/99 at 13:11:06 From: Doctor Rob Subject: Re: Factorization in complex field Thanks for writing to Ask Dr. Math. No, it is not always possible. If it were possible, you could do it with g1 = 1. If you multiply out the product, you will get: (h1)*x^2 + (g2*h2)*y^2 + (g3*h3)*z^2 + (h2+g2*h1)*x*y + (h3+g3*h1)*x*z + (g2*h3+g3*h2)*y*z. For this to equal k1*x^2 + k2*y^2 + k3*z^2 + k4*x*y + k5*x*z + k6*y*z, you need the following system of equations to have a solution: k1 = h1, k2 = g2*h2, k3 = g3*h3, k4 = h2 + g2*h1, k5 = h3 + g3*h1, k6 = g2*h3 + g3*h2. Then (h2-g2*h1)^2 = k4^2 - 4*k1*k2, h2 - g2*h1 = +-sqrt(k4^2-4*k1*k2). h2 = (k4+-sqrt(k4^2-4*k1*k2))/2, g2*h1 = (k4-+sqrt(k4^2-4*k1*k2))/2. Similarly we can solve for h3, g3*h1, g2*h3, and g3*h2, all in terms of k1 through k6. Using these, you can solve for g2 = (g2*h3)/h3, and g3 = (g3*h2)/h2, and then h1 = (g3*h1)/g3, all in terms of k1 through k6. Now you substitute g2 and h1 into the equation: g2*h1 = (k4-+sqrt(k4^2-4*k1*k2))/2 from above, and you'll get a condition on the k's that must be satisfied in order for this to work. Sets of k's that don't satisfy this condition cannot be factored in the way you describe above. Example: Factor x^2 + y^2 + z^2 + A*y*z. k1 = 1, k2 = 1, k3 = 1, k4 = 0, k5 = 0, k6 = A, 1 = h1, 1 = g2*h2, 1 = g3*h3, 0 = h2 + g2*h1, 0 = h3 + g3*h1, A = g2*h3 + g3*h2. h2 = +-i, g2*h1 = -+i, h3 = +-i, g3*h1 = -+i, g2*h3 = (A+-sqrt(A^2-4))/2, g3*h2 = (A-+sqrt(A^2-4))/2, g2 = (g2*h3)/h3, = ((A+-sqrt(A^2-4))/2)*(-+i), g3 = (g3*h2)/h2, = ((A-+sqrt(A^2-4))/2)/(+-i), h1 = (-+i)/(((A-+sqrt(A^2-4))/2)/(+-i)), = (A+-sqrt(A^2-4))/2, -+i = ((A+-sqrt(A^2-4))/2)*(-+i)*(A+-sqrt(A^2-4))/2, 4 = A^2 +-2*A*sqrt(A^2-4) + A^2 - 4, 4 - A^2 = +-A*sqrt(A^2-4), 16 - 8*A^2 + A^4 = A^4 - 4*A^2, 4 = A^2, A = +-2. Those two values will allow factorization of the original polynomial, and any other will not. In particular, if we set A = 0, we'll see that x^2 + y^2 + z^2 cannot be so factored. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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