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In approximate arithmetic, such as floating-point arithmetic, the distributive property of multiplication (and division) over addition may fail because of the limitations of arithmetic precision. For example, the identity 1 / 3 + 1 / 3 + 1 / 3 = ( 1 + 1 + 1 ) / 3 {\displaystyle 1/3+1/3+1/3=(1+1+1)/3} fails in decimal arithmetic , regardless of ...
The FOIL method is a special case of a more general method for multiplying algebraic expressions using the distributive law. The word FOIL was originally intended solely as a mnemonic for high-school students learning algebra. The term appears in William Betz's 1929 text Algebra for Today, where he states: [2]
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
The formula for the difference of two squares can be used for factoring polynomials that contain the square of a first quantity minus the square of a second quantity. For example, the polynomial x 4 − 1 {\displaystyle x^{4}-1} can be factored as follows:
When the characteristic of K is 2, so that 2 is not a unit, it is still possible to use a quadratic form to define a symmetric bilinear form B′(x, y) = Q(x + y) − Q(x) − Q(y). However, Q ( x ) can no longer be recovered from this B ′ in the same way, since B ′( x , x ) = 0 for all x (and is thus alternating). [ 8 ]
A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial: x 6 − 9 x 3 + 8 = 0. {\displaystyle x^{6}-9x^{3}+8=0.} Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem ).
Here we plot the Chebyshev nodes of the first kind and the second kind, both for n = 8. For both kinds of nodes, we first plot the points equi-distant on the upper half unit circle in blue. Then the blue points are projected down to the x-axis. The projected points, in red, are the Chebyshev nodes.
One way to see this is to note that the graph of the function f(x) = x 2 is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function f(x − h) = (x − h) 2 is a parabola shifted to the right by h whose vertex is at (h, 0), as shown in the top figure.