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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 ...
Since x 2 represents the area of a square with side of length x, and bx represents the area of a rectangle with sides b and x, the process of completing the square can be viewed as visual manipulation of rectangles. Simple attempts to combine the x 2 and the bx rectangles into a larger square
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...
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).
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 ...
Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression: 1 − x 2 1 + x 2 {\displaystyle {\sqrt {\frac {1-x^{2}}{1+x^{2}}}}} An algebraic equation is an equation involving polynomials , for which algebraic expressions may be solutions .
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.
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 ...