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  • ...cases where <math>k < n</math>, an argument based on analyzing individual monomials in the expansion can be used (see http://web.stanford.edu/~marykw/classes/C
    4 KB (704 words) - 07:28, 24 November 2024
  • ...[[equation]]s in which both sides of the equation are [[polynomials]] or [[monomials]] of the first [[degree]] - i.e. each term does not have any variables to a
    1 KB (257 words) - 12:39, 14 July 2021
  • ...triangle to determine the lengths <math>YA'</math> and <math>YM'</math> as monomials in <math>x</math>. Thus, the locus of the midpoint will be linear between e
    4 KB (707 words) - 15:36, 15 February 2021
  • ...th>x^9-x</math> is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
    782 bytes (120 words) - 21:14, 3 October 2014
  • ...th>x^9-x</math> is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
    17 KB (2,459 words) - 21:40, 10 April 2023
  • We first prove the result for monomials, polynomials in which only one coefficient is nonzero. This is obvious for ...the terms not interfering with each other. If <math>f</math> contains any monomials that do not satisfy the conditions of the problem, then picking suitable <m
    4 KB (735 words) - 15:08, 18 March 2023
  • ...ts, x_n)</math> cannot be written as the sum of fewer than <math>n!</math> monomials. (A monomial is a polynomial of the form <math>cx^{d_1}_1 x^{d_2}_2\ldots x
    776 bytes (138 words) - 18:16, 6 October 2023
  • ...ts, x_n)</math> cannot be written as the sum of fewer than <math>n!</math> monomials. (A monomial is a polynomial of the form <math>cx^{d_1}_1 x^{d_2}_2\ldots x
    5 KB (799 words) - 15:46, 5 August 2023
  • <math>g(x)</math> is a sum of these monomials so this gives us a method to determine the sum we're looking for:
    8 KB (1,183 words) - 04:40, 26 July 2024
  • ...integer <math>k</math>, assign a monomial <math>x^k</math>. The sum of the monomials is <cmath> x(1+x+\ldots+x^{n^2-1}) = \sum_{i=1}^n x^{a_i}(1+x^{d_i}+\ldots+
    3 KB (656 words) - 20:07, 11 February 2024