(7/x^2+x-12)+(3/x^2+16) I know it factors to: (7/(x+4)(x-3))+(3/(x+4)(x-4))
B bmw1 New member Joined Apr 9, 2014 Messages 1 Apr 9, 2014 #1 (7/x^2+x-12)+(3/x^2+16) I know it factors to: (7/(x+4)(x-3))+(3/(x+4)(x-4)) Last edited: Apr 9, 2014
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,187 Apr 9, 2014 #2 bmw1 said: (7/x^2+x-12)+(3/x^2+16) I know it factors to: (7/(x+4)(x-3))+(3/(x+4)(x-4)) Click to expand... You wrote those fractions wrong, because you're missing grouping symbols around the denominators. 7/(x^2 + x - 12) + 3/(x^2 + 16) But x^2 + 16 is prime, so check to see if it is supposed to be "x^2 - 16." I suspect it is. If that is the case, then you actually have: 7/(x^2 + x - 12) + 3/(x^2 - 16) = 7/[(x + 4)(x - 3)] \(\displaystyle \ \) + \(\displaystyle \ \) 3/[(x + 4)(x - 4)]
bmw1 said: (7/x^2+x-12)+(3/x^2+16) I know it factors to: (7/(x+4)(x-3))+(3/(x+4)(x-4)) Click to expand... You wrote those fractions wrong, because you're missing grouping symbols around the denominators. 7/(x^2 + x - 12) + 3/(x^2 + 16) But x^2 + 16 is prime, so check to see if it is supposed to be "x^2 - 16." I suspect it is. If that is the case, then you actually have: 7/(x^2 + x - 12) + 3/(x^2 - 16) = 7/[(x + 4)(x - 3)] \(\displaystyle \ \) + \(\displaystyle \ \) 3/[(x + 4)(x - 4)]
