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For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorization is not usually considered meaningful within number systems possessing division , such as the real or complex numbers , since any x {\displaystyle x} can be trivially written as ( x y ) × ( 1 / y ) {\displaystyle ...
[17] [18] For example, the fraction 1/(x 2 + 1) is not a polynomial, and it cannot be written as a finite sum of powers of the variable x. For polynomials in one variable, there is a notion of Euclidean division of polynomials, generalizing the Euclidean division of integers.
If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as 1×2, 2×1, (−1)×(−2), or (−2)×(−1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values p(0) = 1, 2, −1, or −2. and likewise for p(1).
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
Moreover, if one sets x = 1 + t, one gets without computation that () = (+) is a polynomial in t with the same first coefficient 3 and constant term 1. [2] The rational root theorem implies thus that a rational root of Q must belong to { ± 1 , ± 1 3 } , {\textstyle \{\pm 1,\pm {\frac {1}{3}}\},} and thus that the rational roots of P satisfy x ...
[1] [2] [3] [better source needed]. For example, 3 x 2 − 2 x y + c {\displaystyle 3x^{2}-2xy+c} is an algebraic expression. Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression:
The polynomial P = x 4 + 1 is irreducible over Q but not over any finite field. On any field extension of F 2, P = (x + 1) 4. On every other finite field, at least one of −1, 2 and −2 is a square, because the product of two non-squares is a square and so we have; If =, then = (+) ().
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...