Search results
Results From The WOW.Com Content Network
In mathematics, factorization (or factorisation, see English spelling differences) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. For example, 3 × 5 is an integer factorization of 15, and (x – 2) (x + 2) is a polynomial ...
Irreducible polynomial. In mathematics, an irreducible polynomial is, roughly speaking, a polynomial that cannot be factored into the product of two non-constant polynomials. The property of irreducibility depends on the nature of the coefficients that are accepted for the possible factors, that is, the ring to which the coefficients of the ...
The graph of a degree 1 polynomial (or linear function) f(x) = a0 + a1x, where a1 ≠ 0, is an oblique line with y-intercept a0 and slope a1 . The graph of a degree 2 polynomial. f(x) = a0 + a1x + a2x2, where a2 ≠ 0. is a parabola . The graph of a degree 3 polynomial. f(x) = a0 + a1x + a2x2 + a3x3, where a3 ≠ 0.
Partial fraction decomposition. 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 ...
Polynomial expansion. In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by ...
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1. [12]
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).