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[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 ...
For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorization 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 ) {\displaystyle ...
If two or more factors of a polynomial are identical, then the polynomial is a multiple of the square of this factor. The multiple factor is also a factor of the polynomial's derivative (with respect to any of the variables, if several). For univariate polynomials, multiple factors are equivalent to multiple roots (over a suitable extension field).
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
Given a quadratic polynomial of the form + the numbers h and k may be interpreted as the Cartesian coordinates of the vertex (or stationary point) of the parabola. That is, h is the x -coordinate of the axis of symmetry (i.e. the axis of symmetry has equation x = h ), and k is the minimum value (or maximum value, if a < 0) of the quadratic ...
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 ...
The Barth surface, shown in the figure is the geometric representation of the solutions of a polynomial system reduced to a single equation of degree 6 in 3 variables. Some of its numerous singular points are visible on the image. They are the solutions of a system of 4 equations of degree 5 in 3 variables.
Divide the first term of the dividend by the highest term of the divisor (x 3 ÷ x = x 2). Place the result below the bar. x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by ...