and so why is this??? x(sqrd)/x+2 - 4/x+2?

Question

I hate fractions.

Answer 1

Presumably you mean
(x^2)/(x + 2) - 4(/x + 2)
Since the denominators are the same, you can add the numerators without modifying them:
(x^2 - 4)/(x + 2)
Factor the numerator to get
(x - 2)(x + 2)/x+2)
(x + 2)/x+2) = 1 so you are left with
(x - 2)

If you post this question under philosophy, you can probably get many different answers to why this is. Because it works is mine.

Answer 2

These fractions have the smae denominator so it's just like subtracting 5/7-3/7. You get 4/7

Now in algebra, as in arithmetic, you have to reduce if possible. To do that you must factor and then divide out the common factors. Here goes.

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

(x+2)(x-2)/(x+2)

x-2 is the answer

Answer 3

Problem: [(x^2)/(x+2)] - [4/(x+2)]

You first need to combine the fractions.

Since they both have the same denominator, we can simply write the equation as:

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

You now need to factor x^2 - 4.

You may need to think of x^2 - 4 as x^2 + 0x - 4.

Ask yourself what two numbers multiply to be -4 and add to be 0.

You will find that 2 and -2 multiply to be -4 and add to be 0.

So x^2 - 4 factors into (x - 2)(x + 2)

Your equation now becomes [(x-2)(x+2)] / (x+2)

The (x+2) in the denominator and numerator will cancel to make 1 leaving you with

(x - 2) as your answer.

Hope this helps.

Answer 4

taking x+2 as the common denominator
x^2-4/x+2
factoring the Nr
(x+2)(x-2)/(x+2)
canceling x+2 and reducing the fraction to the simplest form
=(x-2) is the answer

Answer 5

x² / x + 2 - 4 / x + 2

x² - 4 = x + 2)(x - 2)/ x + 2. . .X + 2 cancell

The answer is (x - 2)

Answer 6

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

because a^2-b^2 = (a-b)(a+b)

Answer 7

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

Answer 8

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

i hope that this helps

Answer 9

x-2 is the awnser but your domain is still (-infinity,-2)U(-2,infinity)