distributive
(a-b) (a+b) = a2+b2
multiply the 1st term with whole bracket and the 2nd term with whole bracket
no, because some examples are: (a-2)(a+2) = a^2-4 (binomial) & (a+b)(c-d) = ac-ad+bc-db (polynomial) but can 2 binomials equal to a monomial?
The ones that are the sum or the difference of two terms.
a²-b²
the two consecutive positive integers whose product is 380 19 20
distributive.
distributive
(a-b) (a+b) = a2+b2
no please give me 5 riddles about product of 2 binomial
multiply the 1st term with whole bracket and the 2nd term with whole bracket
It means that the question has been written by someone who does not know what the word "polynomial" means, or else that this is a copy-and-paste by someone who knows even less! Only a trinomial can be written as a product of two binomials. No polynomial of any other order can!
Binomials are algebraic expressions of the sum or difference of two terms. Some binomials can be broken down into factors. One example of this is the "difference between two squares" where the binomial a2 - b2 can be factored into (a - b)(a + b)
no, because some examples are: (a-2)(a+2) = a^2-4 (binomial) & (a+b)(c-d) = ac-ad+bc-db (polynomial) but can 2 binomials equal to a monomial?
No. A counter-example proves the falsity: Consider the two binomials (x + 2) and (x - 2). Then (x + 2)(x - 2) = x2 - 2x + 2x - 4 = x2 - 4 another binomial.
The two binomials can be written as (x - a)(x + a), for any constant a. Proof: Expand using FOIL: (x - a)(x + a) = x2 + xa - xa - a2 Group: (x - a)(x + a) = x2 - a2 x2 - a2 is a difference of squares. Thus, the product of (x - a) and (x + a) is a difference of squares.