How do you write #(5n)^2# without exponents?

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Answer 1 · 2017-02-20T23:11:37

#(5n)^2 = (5n)(5n) = (5*5)(n*n) = 25n * n#

#(5n)^2 = (5n)(5n) = (5*5)(n*n) = 25n * n#

Answer 2 · 2017-02-20T23:13:04

#(5n)^2 = 25*n*n#

For any positive integer #m# and number #a# we have:

#a^m = overbrace(a*a*...*a)^"m times"#

In particular:

#a^2 = a*a#

Putting #a=5n# we find:

#(5n)^2 = (5n)*(5n) = 5*n*5*n = 5*5*n*n = 25*n*n#