(n + 8)(n - 8)
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100*n/64100*n/64100*n/64100*n/64
Suppose you have a number N and you want to find its largest prime factor. It is probably easiest to start at the bottom.Find the smallest prime factor, p.Find its factor pair = N/p.If the second number (= N/p) is a prime, then it is the largest prime factor.if not, replace N by N/p and go back to the top.
Did you mean find n where (2^5)^n = 64? (Sorry - the superscript symbol won't work!) I think that would need logarithms to calculate, because 2^5 = 32. Or did you mean find the power of 2 that = 64?
n = number64/n====
Yes. First you need to factor out a 3 so the equation becomes 3(n2-8n+12). Then you need to factor what's in the parenthesis, which would be (n-6)(n-2). So, the answer is 3(n-6)(n-2).