How do you simplify and state the excluded values for #(y+5) /( y^2+4y-32)#?

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Answer 1 · 2015-06-03T22:48:56

#(y+5)/(y^2+4y-32) = (y+5)/((y+8)(y-4))#

If you like, you can eliminate #y# from the numerator, by writing:

#(y+5)/((y+8)(y-4)) =((y-4) + 9)/((y+8)(y-4))#

#=1/(y+8)+9/((y+8)(y-4))#

or by writing:

#(y+5)/((y+8)(y-4)) =((y+8) - 3)/((y+8)(y-4))#

#=1/(y-4) - 3/((y+8)(y-4))#

In any case, the excluded values are #y=4# and #y=-8#