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
1,3 a b c
That factors to (a + 1)(a + b) a = -1, -b b = -a