How do you simplify and find the restrictions for #( (x^2 -x-12)/( x^2 -2x-15) ) / ( (x^2 + 8x+12)/( x^2 -5x-14))#?

1 Answer
Oct 5, 2016

Answer:

The expression simplifies to #((x - 4)(x + 7))/((x - 5)(x + 6))# with restrictions being #x!= 5, -3, -6, 7 and -2#.

Explanation:

#=(((x - 4)(x + 3))/((x - 5)(x + 3)))/(((x + 6)(x + 2))/((x - 7)(x + 2))#

#=((x - 4)/(x - 5))/((x + 6)/(x + 7))#

#=((x -4)(x + 7))/((x - 5)(x + 6))#

Restrictions occur when the denominator of the original expression equal #0#. We can determine restrictions by setting the denominator to #0# and solving.

#(x - 5)(x + 3) = 0#

#x = 5 and -3#

AND

#(x + 6)(x + 2) = 0#

#x = -6 and -2#

AND

#(x - 7)(x + 2) = 0#

#x = 7 and -2#

Hopefully this helps!