How do you simplify #(8x^5-10xy^2)/(2xy^2)# and what are the excluded values of the variables?

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Answer 1 · 2017-04-18T06:45:26

#=(4x^4-5y^2)/y^2" or "=(4x^4)/y^2 -5#

#xand y# may not be #0#

The excluded values are any values that would make the denominator equal to #0#, or any number that we use to divide.
#xand y# may not be #0#

#(8x^5 -10xy^2)/(2xy^2)" "larr# factorise the numerator

#=(2x(4x^4-5y^2))/(2xy^2)#

#=(cancel(2x)(4x^4-5y^2))/(cancel(2x)y^2)#

#=(4x^4-5y^2)/y^2#

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You could also split the fraction into two separate fractions:

#(8x^5)/(2xy^2) -(10xy^2)/(2xy^2)#

#=(4x^4)/y^2 -5#

This leaves two terms in the answer.