The problem is :

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

^2=squared

ive been on a lot of math help websites but I cant seem to figure out how to solve this

The problem is :

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

^2=squared

ive been on a lot of math help websites but I cant seem to figure out how to solve this

Tutors, please sign in to answer this question.

Simplify first by factoring,

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

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

= 3/(x+1) + 1/(x+1)

= 4/(x+1), where x ≠ 1.

Daisy:

It seems like your other responders are using a different starting expression than I am reading in your post (?). Correct me if I'm wrong, but the expression you have written (I believe) is as follows:

**(3x+3/x ^{2}+2x+1)+(x-1/x^{2}-1)**

I would start by removing the parentheses and combining like terms to get **
3x+2x+x+3/x ^{2}-1/x^{2}+1-1**

Next simplify to get **6x+2/x ^{2}**

You can then multiply the first term by x^{2}/x^{2} to get the two remaining terms over the same denominator:
**6x ^{3}/x^{2}+2/x^{2}**

Now add the numerators to get the final result of **(6x ^{3}+2)/x^{2}**

**Again, if my read on your expression is correct!**

**George T.
**

Just as with adding fractions such as 1/4 and 1/3, we must have a common denominator. The least common multiple of 3 and 4 is 12. You need the least common multiple of x^2 + 2x + 1 and x^2 - 1. To do so we must first see what common factors they have, and in order to do so we must factor each one.

x^2 + 2x + 1 = (x + 1)(x + 1)

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

so the least common denominator is (x + 1)(x + 1)(x - 1) which equals x^3 + x^2 - x - 1

So now you need to change each fraction just as you would with 1/4 + 1/3 = 3/12 + 4/12 = 7/12

You need to multiply the first fraction (both the numerator and denominator each) by (x - 1)/(x -1).

((x-1)/(x-1)) ((3x + 3)/(x^2 +2x + 1)) = (3x^2 + 6x + 3)/(x^3 + x^2 - x - 1) which is our new first team, which has the same value as the original since (x-1)/(x-1) equals one.

Second term needs (x + 1) so: ((x+ 1)/(x + 1)) ((x - 1)/(x^2 - 1) = (x^2 - 1)/(x^3 + x^2 - x - 1).

New addition: (3x^2 + 6x + 3)/(x^3 +x^2 -x -1) + (x^2 - 1)/(x^3 +x^2 -x -1) = (4x^2 + 6x + 2)/(x^3 +x^2 -x -1)

Now, this is our sum...however always check if it can be reduced by factoring both the numerator and denominator as much as possible. The denominator is just (x + 1)(x + 1)(x - 1) that we found earlier.

The numerator has a GCF of 2: 2( 2x^2 + 3x + 1)

Factor by grouping: 2 ( 2x^2 + 2x + x + 1)

2 ( 2x( x + 1) + 1( x + 1))

2 ( 2 x + 1 ) ( x + 1).

The numerator and denominator have a common factor of (x + 1). These will cancel out and we are left with 2 (2x + 1) / (x^2 - 1 ). We can distribute the numerator or leave it as it is, it doesn't matter.

All

I am reading Daisy's expression differently than most of you, based on her placement of the parentheses. See my response above

George T.

Daisy,

I'm not exactly clear what rational expressions you meant to add. I realize getting the question written down and communicated is a bit of a challenge with a keyboard, but it makes our time much more effective. This was the expression that I gathered from your post:

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

Taking into account the order of operations, start by squaring the second term:

(x - 1/x^2 - 1)

(x - 1/x^2 - 1)

---------------

-x + 1/x^2 + 1

- 1/x + 1/x^4 + 1/x^2

+ x^2 - 1/x - x

After collecting like terms:

x^2 - 2x - 2/x + 2/x^2 + 1/x^4 + 1

Add the first term to the now squared second term, and collect like terms:

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

x^2 + 3x - 2/x + 5/x^2 + 1/x^4 + 2

The common denominator among the various terms is x^4, by inspection, so one should select various representations of 1 (i.e. x^4/x^4, x^3/x^3, and so on) in order to produce the desired common denominator for each term and multiply:

(x^4/x^4)(x^2 + 3x + 2) - (x^3/x^3)2/x + (x^2/x^2)5/x^2 + 1/x^4

x^6/x^4 + 3x^5/x^4 + 2x^4/x^4 - 2x^3/x^4 + 5x^2/x^4 + 1/x^4

And finally collect like terms once again:

(x^6 + 3x^5 + 2x^4 - 2x^3 + 5x^2 + 1)/x^4

The resulting expression, as was the initial expression, is undefined for the value x = 0.

Thank you for taking the time to try and answer my question! I really appreciate it (:

Edward B.

STAT/MATH/Actuarial Science/MBA/Econ/Fin. - Ivy League Exp & Prof

New York, NY

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Jeffrey Patrick C.

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Dave M.

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## Comments

x ≠-1, looks like you forgot the negative sign.

Here we talked about equivalence. x ≠ 1 should be the corret.

(3x+3)/(x^2+2x+1) + (x-1)/(x^2-1) = 4/(x+1), where x ≠ 1 because if x = 1,then the left side is undefined, but the right side is well defined.