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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 the equivalent sum of products, continuing until the expression becomes a ...
For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorisation ezzzzzzzzz 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 ...
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).
The first algorithm for polynomial decomposition was published in 1985, [6] though it had been discovered in 1976, [7] and implemented in the Macsyma/Maxima computer algebra system. [8] That algorithm takes exponential time in worst case, but works independently of the characteristic of the underlying field.
Horner's method evaluates a polynomial using repeated bracketing: + + + + + = + (+ (+ (+ + (+)))). This method reduces the number of multiplications and additions to just Horner's method is so common that a computer instruction "multiply–accumulate operation" has been added to many computer processors, which allow doing the addition and multiplication operations in one combined step.
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.
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Divide the highest term of the remainder by the highest term of the divisor (x 2 ÷ x = x). Place the result (+x) below the bar. x 2 has been divided leaving no remainder, and can therefore be marked as used. The result x is then multiplied by the second term in the divisor −3 = −3x. Determine the partial remainder by subtracting 0x − ...