3/x^2 +4x+3 minus 1/x^2 – 9?

Answer 1

Hi,

..3...................1
------------..-..------ =
x²+4x+3........x²-9

..3...................1
--------------..-..------------- =
(x+3)(x+1).....(x-3)(x +3)


Multiply 1st fraction by (x-3) on top and bottom and multiply 2nd fraction by (x+1) on top and bottom.

..3(x-3)...................1(x+1)
--------------------..-..--------------------- =
(x+3)(x+1)(x-3).....(x-3)(x +3)(x+1)

Move subtraction sign into numerator and combine into one fraction over the common denominator. Then simplify the numerator.

..3(x-3)..-..1(x+1)
----------------------- =
(x+3)(x+1)(x-3)

3x.-.9.-.x.-.1
----------------------- =
(x+3)(x+1)(x-3)

2x.-.10
-------------------- <== answer
(x+3)(x+1)(x-3)

Answer 2

3/x^2 +4x+3 minus 1/x^2 – 9?
I assume (correctly I hope) that you meant:
3/(x^2 +4x+3) minus 1/(x^2 – 9). If so, the
3/[(x+3)(x+1)] - 1/[(x-3)(x+3)]
=[3(x-3) -(x+1)]/[(x+3)(x-3)(x+1)]
=[3x-3-x-1] /[(x+3)(x-3)(x+1)]
=(2x-4)/[(x+3)(x-3)(x+1)]

Answer 3

=(3/x^2)+4x+3-(1/x^2)+9
=(2/x^2)+4x+12=2((1/x^2)+2x+6)