How do you add or subtract #(x^2+3x+2)/(x^2-16)+(3x+6)/(x^2-16)#?

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Answer 1 · 2015-05-18T16:39:44

Since both fractions have the same denominator, you can simply add up (or substract) the numerators and keep the denominator unchanged :

#(x^2 + 3x +2)/(x^2-16) + (3x+6)/(x^2-16) = ((x^2 + 3x +2)+(3x+6))/(x^2-16)#

#= (x^2 + 6x + 8)/(x^2 -16)#.

Then, you can factorize it :

#(x^2 + 6x + 8)/(x^2 -16) = ((x+2)(x+4))/((x+4)(x-4)) = (x+2)/(x-4)#.

In the case of a substraction :

#(x^2 + 3x +2)/(x^2-16) - (3x+6)/(x^2-16) = ((x^2 + 3x +2)-(3x+6))/(x^2-16)#

#= (x^2 - 4)/(x^2 -16) = ((x+2)(x-2))/((x+4)(x-4))#.