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Animation showing the use of synthetic division to find the quotient of + + + by .Note that there is no term in , so the fourth column from the right contains a zero.. In algebra, synthetic division is a method for manually performing Euclidean division of polynomials, with less writing and fewer calculations than long division.
It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1. Or x and y can both be treated as unknowns, and then there are many solutions to the equation; a symbolic solution is (x, y) = (a + 1, a), where the variable a may take any value. Instantiating a symbolic solution with specific numbers ...
(,) and (,) are given, 2. (,) is given and () is real on the real axis, 3. only (,) is given, 4. only (,) is given. He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4.
Let us consider a polynomial P(x) of degree less than n(m + 1) with indeterminate coefficients; that is, the coefficients of P(x) are n(m + 1) new variables. Then, by writing the constraints that the interpolating polynomial must satisfy, one gets a system of n(m + 1) linear equations in n(m + 1) unknowns. In general, such a system has exactly ...
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 ...
For example, the system x 3 – 1 = 0, x 2 – 1 = 0 is overdetermined (having two equations but only one unknown), but it is not inconsistent since it has the solution x = 1. A system is underdetermined if the number of equations is lower than the number of the variables.
An example of a more complicated (although small enough to be written here) solution is the unique real root of x 5 − 5x + 12 = 0. Let a = √ 2φ −1, b = √ 2φ, and c = 4 √ 5, where φ = 1+ √ 5 / 2 is the golden ratio. Then the only real solution x = −1.84208... is given by
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 ...