For the term AB, which is A x B, the factors are A and B. And 1. One is a factor of everything.
Theo says that a and b are factors of c is this correct
b. 2x2x5 are prime factors
The GCF of ba and b is b. That factors to b(a - 1)
B, as a variable, can stand for any number. The factor possibilities are infinite.
1,2,3,6,a,b,b
b x b = b2
No, but if you're talking about factors, the result is a product. (a × b) × c = a × (b × c)
Factors: 1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500,1000,2000
I wonder if you mean 56a2b2 ? If so you have 2 x 2 x 2 x 7 x a x a x b x b. If a and/or b are not prime, you will have the factors of those as well.
To find the GCF of each pair of monomial of 8ab³ and 10a²b², we can use the following steps: Write the complete factorization of each monomial, including the constants and the variables with their exponents. 8ab³ = 2 ⋅ 2 ⋅ 2 ⋅ a ⋅ b ⋅ b ⋅ b 10a²b² = 2 ⋅ 5 ⋅ a ⋅ a ⋅ b ⋅ b Identify the common factors in both monomials. These are the factors that appear in both factorizations with the same or lower exponent. The common factors are: 2, a, and b² Multiply the common factors to get the GCF. GCF = 2 ⋅ a ⋅ b² = 2ab²
That factors to (a + 1)(a + b) a = -1, -b b = -a
Commutative: a + b = b + a a × b = b × a Associative: (a + b) + c = a + (b + c) (a × b) × c = a × (b × c) Commutative states that the sum or product remains the same no matter the order of the factors. Associative states that the sum or product remains the same no matter the grouping of the factors.