How do you simplify and restricted value of #(x^2-36)/(3x+6) #?

1 Answer
marfre
Jul 18, 2017

#((x-6)(x+6))/(3(x+2))#; #x != -2#

Explanation:

Given: #(x^2 - 36)/(3x+6)#

The numerator is the difference of squares #(a^2 - b^2) = (a - b)(a + b): (x^2 - 6^2) = (x -6)(x+6)#

Rewriting the given expression yields: #((x -6)(x+6))/(3x + 6)#

The denominator can be factored using the greatest common factor (GCF) of #3: 3(x + 2)#

Rewriting the given expression yields: #((x-6)(x+6))/(3(x+2))#

Since there is a fraction, the denominator cannot be #= 0#. This means that #3(x + 2) != 0#, or #x + 2 != 0#. This occurs when #x = -2#.

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