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If one can put an equation in a factored form E⋅F = 0, then the problem of solving the equation splits into two independent (and generally easier) problems E = 0 and F = 0. When an expression can be factored, the factors are often much simpler, and may thus offer some insight on the problem. For example,
If the polynomial to be factored is + + + +, then all possible linear factors are of the form , where is an integer factor of and is an integer factor of . All possible combinations of integer factors can be tested for validity, and each valid one can be factored out using polynomial long division .
To convert the standard form to factored form, one needs only the quadratic formula to determine the two roots r 1 and r 2. To convert the standard form to vertex form, one needs a process called completing the square. To convert the factored form (or vertex form) to standard form, one needs to multiply, expand and/or distribute the factors.
[6]: 202–207 If one is given a quadratic equation in the form x 2 + bx + c = 0, the sought factorization has the form (x + q)(x + s), and one has to find two numbers q and s that add up to b and whose product is c (this is sometimes called "Vieta's rule" [7] and is related to Vieta's formulas). As an example, x 2 + 5x + 6 factors as (x + 3)(x ...
Every morphism f of C can be factored as = for some morphisms and . The factorization is functorial : if u {\displaystyle u} and v {\displaystyle v} are two morphisms such that v m e = m ′ e ′ u {\displaystyle vme=m'e'u} for some morphisms e , e ′ ∈ E {\displaystyle e,e'\in E} and m , m ′ ∈ M {\displaystyle m,m'\in M} , then there ...
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Construct an ambiguous form (a, b, c) that is an element f ∈ G Δ of order dividing 2 to obtain a coprime factorization of the largest odd divisor of Δ in which Δ = −4ac or Δ = a(a − 4c) or Δ = (b − 2a)(b + 2a). If the ambiguous form provides a factorization of n then stop, otherwise find another ambiguous form until the ...
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