can someone help me simplify this?

Question

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

apparently, i'm really bad at simplifying...

Answer 1

x^2 - 3x + 2 = (x-2)*(x-1)

(x-4)/(x^2 - 3x + 2) = (x-4)/((x-2)*(x-1))

(x^2-4) = (x-2)*(x+2)
Therefore:
(x+3)/(x^2-4) = (x+3)/((x-2)*(x+2))

So now we have:

(x-4)/((x-2)*(x-1)) + (x+3)/((x-2)*(x+2))

The denominators are not the same so we can try to find the Greatest Common Denominator (GCD). After finding that, everything will be a piece of cake. I think you can do that by yourself.

Good luck.

Answer 2

Cheer up, alot of people are. It's a lot of grind.
The first denominator factors to (x-2)(x-1) and the second to (x-2)(x+2). So the LCD is (x-2)(x-1)(x+2). Doing the math:
[(x-4)*(x+2) + (x+3)*(x-1)]/ [LCD]

now you're going to have to multiply the factors togeher in the numerator and combine like power terms: Thus...
x^2-2x-8 + x^2 +2x -3= 2x^2-11. It look like this won't factor in anything useful, so quit while the going is good. The answer would be
(2x^2-11)/[(x-2)(x-1)(x+2)]
I don't know the rules at your school; they might want you to multiply out the denominator.

Answer 3

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

(x-4)/(x - 2)(x - 1)+(x+3)/(x + 2)(x - 2) ---->Factored.

Now get a common denominator.

PS this is all in one line:
[(x-4)/(x-2)(x-1)*(x+2)/(x+2)] + ...
..[(x+3)/(x+2)(x-2)*(x-1)/(x-1)

[(x-4)(x+2)/(x-2)(x-1)(x+2)] +...
..[(x+3)(x-1)/(x+2)(x-2)(x-1)]

Add together - with common denominator on the bottom.

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

= (x^2 - 2x - 8) + (x^2 + 2x - 3) / (x+2)(x-2)(x-1)
= (2x^2 -11) / (x+2)(x-2)(x-1) ANSWER
or
= (2x^2 - 11) / (x^2- 4)(x-1)
= (2x^2 - 11) / (x^3 - x^2 - 4x + 4) ANSWER

Answer 4

(x-4)/(x^2-3x+2)+(x+3)/(x^2-4)
= (x-4)/[(x-1)(x-2)] + (x+3)/[(x+2)(x-2)]
= [(x-4)(x+2) + (x+3)(x-1)] / [(x-1)(x+2)(x-2)]
= (x^2 - 2x - 8 + x^2 + 2x - 3) / (x^3 - x^2 - 4x + 4)
= (2x^2 - 11) / (x^3 - x^2 - 4x + 4)