How do you simplify #(64-c^2)/(c^2-7c-8)# and what are the ecluded values fot he variables?

Question

1354 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-16T21:00:25

this expression can be simplified to #-(c + 8)/(c + 1)# with restrictions of #c!= 8# and #c!=-1#.

Factor.

#=-(c^2 - 64)/(c^2 - 7c - 8)#

#=-((c + 8)(c - 8))/((c - 8)(c + 1))#

#=-(c + 8)/(c + 1)#

As for the excluded values, these are found by setting the denominator of the initial expression to #0# and solving for the variable.

#c^2 - 7c - 8 = 0#

#(c - 8)(c + 1)#

#c = 8 and -1#

Hence, #c!= 8# and #c!=-1#.

Hopefully this helps!