How do you simplify #(b^2+2b-24)/(2b^2-72)# and find any non permissible values?

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Answer 1 · 2016-10-23T22:27:34

The expression simplifies to #(b - 4)/(2(b - 6))#, #b !=-6, 6#.

Factor everything.

#=((b + 6)(b - 4))/(2(b^2 - 36))#

#=((b + 6)(b - 4))/(2(b + 6)(b - 6))#

#=(b - 4)/(2(b - 6))#

For the non-permissible values, these will occur when the denominator equals #0#.

#0 = 2b^2 - 72#

#0 = 2(b^2 - 36)#

#0 = (b + 6)(b - 6)#

#b = -6 and 6#

Hopefully this helps!