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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, + + + having 16 multiplications, 4 ...
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 elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form + + to the form + for some values of and . [1] In terms of a new quantity x − h {\displaystyle x-h} , this expression is a quadratic polynomial with no linear term.
Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
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.
has no real number solution since no real number squared equals −1. Sometimes a quadratic equation has a root of multiplicity 2, such as: (+) = For this equation, −1 is a root of multiplicity 2. This means −1 appears twice, since the equation can be rewritten in factored form as
For example, antiderivatives of x 2 + 1 have the form 1 / 3 x 3 + x + c. For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p , or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ...
If an equation P(x) = 0 of degree n has a rational root α, the associated polynomial can be factored to give the form P(X) = (X – α)Q(X) (by dividing P(X) by X – α or by writing P(X) – P(α) as a linear combination of terms of the form X k – α k, and factoring out X – α. Solving P(x) = 0 thus reduces to solving the degree n – 1 ...