How do you add or subtract (x^2+3x+2)/(x^2-16)+(3x+6)/(x^2-16)x2+3x+2x216+3x+6x216?

1 Answer
May 18, 2015

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)x2+3x+2x216+3x+6x216=(x2+3x+2)+(3x+6)x216

= (x^2 + 6x + 8)/(x^2 -16)=x2+6x+8x216.

Then, you can factorize it :

(x^2 + 6x + 8)/(x^2 -16) = ((x+2)(x+4))/((x+4)(x-4)) = (x+2)/(x-4)x2+6x+8x216=(x+2)(x+4)(x+4)(x4)=x+2x4.

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)x2+3x+2x2163x+6x216=(x2+3x+2)(3x+6)x216

= (x^2 - 4)/(x^2 -16) = ((x+2)(x-2))/((x+4)(x-4))=x24x216=(x+2)(x2)(x+4)(x4).