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For instance, the polynomial x 2 + 3x + 2 is an example of this type of trinomial with n = 1. The solution a 1 = −2 and a 2 = −1 of the above system gives the trinomial factorization: x 2 + 3x + 2 = (x + a 1)(x + a 2) = (x + 2)(x + 1). The same result can be provided by Ruffini's rule, but with a more complex and time-consuming process.
Solving these two quintics yields r = 1.501 × 10 9 m for L 2 and r = 1.491 × 10 9 m for L 1. The Sun–Earth Lagrangian points L 2 and L 1 are usually given as 1.5 million km from Earth. If the mass of the smaller object ( M E ) is much smaller than the mass of the larger object ( M S ), then the quintic equation can be greatly reduced and L ...
These factorizations work not only over the complex numbers, but also over any field, where either –1, 2 or –2 is a square. In a finite field , the product of two non-squares is a square; this implies that the polynomial x 4 + 1 , {\displaystyle x^{4}+1,} which is irreducible over the integers, is reducible modulo every prime number .
Layers of Pascal's pyramid derived from coefficients in an upside-down ternary plot of the terms in the expansions of the powers of a trinomial – the number of terms is clearly a triangular number. In mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomials. The expansion is given by
In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1).
Solving an equation symbolically means that expressions can be used for representing the solutions. For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement.
Given a quadratic polynomial of the form + + it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: + + = [+ +] = [(+) +] = (+) + = (+) + This process of factoring out the coefficient a can further be simplified by only factorising it out of the first 2 terms.
The first person who conjectured that the problem of solving quintics by radicals might be impossible to solve was Carl Friedrich Gauss, who wrote in 1798 in section 359 of his book Disquisitiones Arithmeticae (which would be published only in 1801) that "there is little doubt that this problem does not so much defy modern methods of analysis ...