How do you write the prime factorization of #28x^2y#?

1 Answer
Alan N.
Jan 14, 2017

#2xx2xx7xx x xx x xx y#
[With the assumption that #x# and #y# are prime.]

Explanation:

The prime factors of 28 are: #2xx2xx7#

Since #x# are #y# are unknowns, I will express the factors of #x^2y# as: #x xx x xx y# [With the assumption that #x# and #y# are prime.]

Hence the prime factorisation of #28x^2y# is: #2xx2xx7xx x xx x xx y#

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