How do you simplify #(32n^2p)/(2n^4p)# and what are the ecluded values fot he variables?

Question

1472 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-30T13:52:28

See the entire solution process below:

First, rewrite this expression as:

#(32/2)(n^2/n^4)(p/p) = 16 xx n^2/n^4 xx 1 = 16(n^2/n^4)#

Now, use this rule of exponents to simplify the #n# terms:

#x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#16(n^color(red)(2)/n^color(blue)(4b)) = 16(1/n^(color(blue)(4)-color(red)(2))) = 16(1/n^2) = 16/n^2#

From the original expression we cannot divide by #0#, therefore the excluded values are:

#2n^4p# cannot equal #0# or #n != 0# and #p != 0#