How to Solve Polynomials Warm-up Facts to know
... 9. The factorizations of x3 ± y 3 on the other side of the page have their generalizations. (a) Similarly to the factorization of x3 − y 3 , you can factor x − y out of xn − y n for any n. (Why?) What is the resulting identity? (b) Similarly to the factorization of x3 + y 3 , you can factor x + y ou ...
... 9. The factorizations of x3 ± y 3 on the other side of the page have their generalizations. (a) Similarly to the factorization of x3 − y 3 , you can factor x − y out of xn − y n for any n. (Why?) What is the resulting identity? (b) Similarly to the factorization of x3 + y 3 , you can factor x + y ou ...
THE HILBERT SCHEME PARAMETERIZING FINITE LENGTH
... (5) The A–module A ⊗k k[x](x) /(F (x)) is free of rank n with a basis consisting of the classes of 1, x, . . . , xn−1 . (6) For all maximal ideals P of A with residue map ϕ: A → κ(P ), the κ(P )– vectorspace κ(P )⊗k k[x](x) /(F ϕ (x)) is n–dimensional with a basis consisting of the classes of 1, x, ...
... (5) The A–module A ⊗k k[x](x) /(F (x)) is free of rank n with a basis consisting of the classes of 1, x, . . . , xn−1 . (6) For all maximal ideals P of A with residue map ϕ: A → κ(P ), the κ(P )– vectorspace κ(P )⊗k k[x](x) /(F ϕ (x)) is n–dimensional with a basis consisting of the classes of 1, x, ...
x - ClassZone
... Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem. 4 + 6 is also a zero by the irrational conjugate theorem. Use the five zeros and the factor theorem to write f(x) as a product of five factors. SOLUTION = (x – 2) [ (x2 –(2i)2][x2–4)+√6 ...
... Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem. 4 + 6 is also a zero by the irrational conjugate theorem. Use the five zeros and the factor theorem to write f(x) as a product of five factors. SOLUTION = (x – 2) [ (x2 –(2i)2][x2–4)+√6 ...
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