3/x^2+4x+3 - 1/x^2-9?

Question

Simpliest form... !! YIKES help please

Answer 1

First factor the denominators by "FOIL"

x^2 + 4x + 3 = (x+3)(x+1); x^2 - 9 = (x+3)(x-3)

The common denominator is (x+3)(x+1)(x-3)

Your first fraction need the (x-3) so multiply top & bottom by (x-3)

The new top is 3(x-3) or 3x - 9

Your second fraction need the (x+1) so multiply top & bottom by (x+1)

Its new top is -1(x+1) or -1x - 1

Now combine like terms on top since they have a common denominator

It will be 2x - 10 which is 2(x-5)

So the answer is 2(x-5) over (x+3)(x+1)(x-3)

Answer 2

3/(x^2 +4x +3) - 1/(x^2-9) =
3/(x+3)(x+1) -1 / (x- 3) (x +3 ) =
1/(x+3) [ 3/(x+1) - 1/(x-3) ] =
1/(x+3) { (3x-9 -x -1) / (x+1 ) (x-3 ) } =
( 2x -10 ) / (x+3 ) (x -3) (x +1) =
2 (x-5) / (x+1) (x+3) (x-3)

Answer 3

3 / (x^2 + 4x +3) - 1 / (x^ - 9)

3 / (x + 3)(x + 1) - 1 / (x + 3)(x - 3)

3(x-3) / (x-3)(x+3)(x+1) - 1(x+1) / (x+1)(x+3)(x-3)

(3x - 9 - x - 1) / (x+3)(x-3)(x+1)

(2x - 10) / (x+3)(x-3)(x+1)

2(x-5) / (x+3)(x-3)(x+1)

Answer 4

3/[x^2+4x+3]-1/[x^2-9]
=3/[x+1][x+3]-1/[x+3][x-3]
=1/[x+3]{3/[x+1]-1/[x-3]}
=1/[x+3][3x-9-x-1]
=1/[x+3]*2[x-4]
=1/2[x+3][x-4]