120 = 2 * 2 * 2 * 3 * 5 Written as a product of powers is (2^3) * 3 * 5
It is: 23*31*51 = 120
As a product of its prime factors: 2*2*2*3*5 = 120
The prime factors of 120 are: 2, 3, 5
The prime factors of 120 are: 2, 3 and 5. 2 x 2 x 2 x 3 x 5 = 120
120 = 23 x 3 x 5
It is: 23*31*51 = 120
As a product of its prime factors: 2*2*2*3*5 = 120
The prime factorization of 120 is not 2345 because 2345 is not a factor of 120. The prime factorization of 120 is 2^3 * 3 * 5, which means that 120 can be expressed as the product of primes 2, 3, and 5.
The prime factors of 120 are 2, 3 and 5.
prime factor, 5
As a product of its prime factors: 23*3*5 = 120
As a product of its prime factors: 2*2*2*3*5 = 120 or as 23*3*5 = 120
As a product of its prime factors: 2*2*2*3*5 = 120 or as 23*3*5 = 120
120 60,2 30,2,2 15,2,2,2 5,3,2,2,2
Start by listing out all pairs of numbers greater than 0 and less than or equal to 120 that have a product equal to 120:120 = 1 x 120120 = 2 x 60120 = 3 x 40120 = 4 x 30120 = 5 x 24120 = 6 x 20120 = 8 x 15120 = 10 x 12So, there are 8 factor pairs of 120. Now, show the prime factorization for 120 by expressing it as the product of ALL prime numbers:120 = 2 x 60
Five
As a product of its prime factors: 2*2*2*3*5 = 120