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How do you simplify #(x^2 + 5x + 4 )/ (x^2 - 16)#?

1 Answer
Mar 9, 2018

Answer:

#(x+1)/(x-4)," ""if "x+4!= 0#

Explanation:

Step 1: Factor both the numerator and denominator.

#(x^2 + 5x + 4)/(x^2 - 16) = [(x+1)(x+4)]/[(x-4)(x+4)]#

Step 2: Cancel common factors.

#[(x+1)cancel((x+4))]/[(x-4)cancel((x+4))]= (x+1)/(x-4)," ""if "x+4!= 0#

Why write this "if #x+4!=0#" part? Because in the original quotient, if #x+4=0# (that is, if #x=–4#), then the denominator would equal zero, and division by zero is undefined.

On its own, #(x+1)/(x-4)# does not give the same problem when #x=–4.# Since we want our simplified expression to reflect the original one in its entirety, we make sure it has all the same restrictions, including #x+4!=0.#