Adding rational expressions. I keep getting these wrong.
x/x^2+6x+9 + 7/x^2+8x+15

- anonymous

Adding rational expressions. I keep getting these wrong.
x/x^2+6x+9 + 7/x^2+8x+15

- Stacey Warren - Expert brainly.com

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- chestercat

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- Nnesha

\[\frac{ x }{ x^2+6x+7 } +\frac{ 7 }{ x^2+8x+15 }\] like this right ?

- anonymous

that was dauntingly fast !

- anonymous

x^2+6x+9 on the first one

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## More answers

- anonymous

5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again.

- anonymous

I've been doing something wrong with factoring on these. it's the correct answer but i'm assuming they want the full answer, not factored. but I have no clue

- Nnesha

haha
alright try to factor quadratic equations apply the AC (headphone method)
find two number when you multiply them you should get product of AC and when you add or subtract them you should get middle term
Ax^2+Bx+C=0
a=leading coefficient
b=middle term
c=constant ter

- Nnesha

okay gimme a sec :)

- Nnesha

so so what are the factors of x^2+6x+9 ??

- Nnesha

so so what are the factors of x^2+8x+15??

- anonymous

I have several of these to answer. will you help me with a few of them? I'm suppose to get 8 out of 10 correct and I was suppose to have all of this finished 22 minutes ago and I don't know if I submit this after i'm finish if it will even count. so i'm kinda panicked. haha I've done this test more than 10 times in the past week and still haven't figured out what i was doing wrong. decided to pay for owlbucks and have someone help me considering the deadline has already passed.

- Nnesha

hmm okay good :)

- Nnesha

do you know how to find factors quadratic equations ?

- Nnesha

* of

- anonymous

No, i don't

- Nnesha

alright
here is the easy method AC method (aka headphone method )
where you have to find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get middle term |dw:1439411259059:dw|
a= leading coefficient which is one
b=middle term = 8
c=constant term = 15
so find two number when you multiply them you should get 15 but when you add or subtract them you should get 8

- anonymous

I solved it with factoring and it was correct.

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @21pmakesmehappy
5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again.
\(\color{blue}{\text{End of Quote}}\)
is it like this ? \[\huge\rm \frac{ 5(x+3)(x-1) }{ (x-3)^2(x+1) }\]
variable is x not y

- anonymous

I redid it and got something different

- anonymous

I have another question now. Can we try this one?

- Nnesha

alright! :) post

- Nnesha

well answer for that one is incorrect
we should try it again!

- anonymous

2y/y^2 -2y+1 + 8/y^2+1y-2

- anonymous

I got the previous question correct once i redid it.

- Nnesha

ohh okay.

- Nnesha

have you tried to solve the 2nd one ?

- anonymous

yes. i'm getting 2(y^2+6y-4)/(y-1)^2(y+2)

- anonymous

but I know it'll be wrong if I put it as my answer.

- anonymous

is it not factored?

- Nnesha

\[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\]
find factor of quadratic equation
y^2-2y+1 = find two number when we multiply them should get 1 and when we add them we should get -2
so -1 times -1= 1
-1-1 = -2 perfect
(x-1)(x-1) are the factors of y^2-2y+1 equation

- Nnesha

it's factorable

- anonymous

i think i need to put in the unfactored answer

- anonymous

because anytime i put it in factored, it's wrong

- Nnesha

1st) factor quadratic equation
2nd) find common denominator

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
\[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\]
find factor of quadratic equation
y^2-2y+1 = find two number when we multiply them should get 1 and when we add them we should get -2
so -1 times -1= 1
-1-1 = -2 perfect
(x-1)(x-1) are the factors of y^2-2y+1 equation
\(\color{blue}{\text{End of Quote}}\)
correction:
(y-1)(y-1)

- Nnesha

\[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\]
\[\rm \frac{ 2y }{ \color{reD}{(y-1)}(y-1) } +\frac{ 8 }{ (y+2)\color{ReD}{(y-1)} }\]
(y+2) (y-1) are factors of y^2+y-2
now what is the common denominator ?

- anonymous

(y-1)^2(y+2)

- anonymous

i have no clue what im doing

- Nnesha

yep right \[\huge\rm \frac{ 2y(y+2) +8(y-1)}{ (y-1)^2(y+2) }\]
multiply the first fraction with the bottom of 2nd fraction
multiply the 2nd fraction with the bottom of first one
and f you distribute parentheses by outside terms you will get \[\frac{ 2y^2+12y-8 }{ (y-1)^2(y+2) }\] which is same as ur answer

- anonymous

it says it was incorrect.

- Nnesha

\[\rm \frac{ 2y^2+12y-8 }{ (y-1)^2(y+2) } \] =\[\frac{ 2(y^2+6y-4) }{ (y-1)^2(y+2)}\]
take out the GCF(greatest common factor

- anonymous

4x/x^2+2x+1 + 3/x^2+3x+2 is the new question

- anonymous

4x^2 +11x+3/(x+1)^2(x+2 is the answer im getting

- anonymous

and i did not. i submitted my answer before you said to take out the gcf so it was incorrect

- anonymous

1 is the gcf, right?

- anonymous

im confused

- anonymous

so do I take 2 out of my final answer?

- Nnesha

no..
GCF is greatest common factor 2 is common in 2y^2 +12y -8
\[\color{ReD}{2} \times 1 \]
\[1 \times \color{Red}{2} \times 3 \times 6 \times 12\] factors of 12
\[1 \times \color{ReD}{2} \times 4 \times 8\] factors of 8
2 is common factor of all 3 terms
so yes take out 2

- anonymous

4x^2+11x+3/(x+1)^2x

- anonymous

is that then?

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
no..
GCF is greatest common factor 2 is common in 2y^2 +12y -8
\[\color{ReD}{2} \times 1 \]
\[1 \times \color{Red}{2} \times 3 \times 6 \times 12\] factors of 12
\[1 \times \color{ReD}{2} \times 4 \times 8\] factors of 8
2 is common factor of all 3 terms
so yes take out 2
\(\color{blue}{\text{End of Quote}}\)
this is for 2nd question!

- anonymous

i am completely lost

- Nnesha

please type the number front of the new question

- Nnesha

like 2nd ) question
3rd) question

- anonymous

i have no clue what question im on

- anonymous

im giving up.

- anonymous

never give up, take a breath get a drink

- anonymous

thanks for your help, though. i appreciate it a lot.

- Nnesha

relax! you are doing a great job!

- Nnesha

alright i didn't give u a GREAT explanation so yeah i'll tag other users

- Nnesha

let's see..

- anonymous

@triciaal 's great

- Nnesha

@Hero @triciaal @nincompoop

- anonymous

is the problem factoring ?

- anonymous

the only reason I have these answers is because I've done them each seriously 10 times today. I know how to solve these problems. I don't need an explanation as to how to solve these problems. i want to know why we match answers and when i put it into the system, it's wrong. I've gotten help from several people over the past week and i can't figure out why my answers are wrong. I appreciate the help but there is honestly no point in attempting this anymore. I've taken this test so many times and the deadline was over an hour ago. I could get all 10 correct and submit it and have it not even count. Your explanation was great and I genuinely appreciate the help. The problem is i get to my final answer, i enter it, it's wrong. I assume it has something to do with the factoring. someone suggested that I give "the full answer" "unfactored" but i have not one clue what that means.

- Nnesha

i would like to see the screenshot of *multiple choice or answer box
what is the statement ??

- anonymous

okay. how can i enter a photo?

- Nnesha

Press PrntScr to take a screenshot
and do you see to blue *attach file* button to attach file ?

- anonymous

I don't see an attach file button

- anonymous

##### 1 Attachment

- Nnesha

http://prntscr.com/83xios

- Nnesha

answer is not correct

- Nnesha

show your work so i can loook over and find out some mistakes
that would be great

- anonymous

This is the last question and i've already gotten 3 wrong. i can go back and take the test after i finish (if im not locked out) so hold on.

- triciaal

do you still need help?

- anonymous

I got locked out. My account is no longer active.

- anonymous

besides factoring the nominator i don't see anything wrong (i'm might be doing it wrong too lol) what exactly is wrong @Nnesha

- Nnesha

okay
like you said you already know how to solve so it's better to post the work
then no one would waste your time

- Nnesha

ik numerator isn't correct so that's mean answer is not correct

- anonymous

well the numerator is correct (again i might be wrong ) just not factored

- triciaal

|dw:1439414757668:dw|

- Nnesha

alright
let's see \[ \frac{ 6a }{ a^2+2a+1 } +\frac{ 7 }{ a^2+3a+2 }\]

- Nnesha

ohhhhhh i see
*facepalm* it's 2 not 7

- triciaal

so your denominator is wrong

- Nnesha

was looking at the wrong question.

- anonymous

im pretty sure LCD is (a+2)(a+1)^2

- Nnesha

ye answer is correct
i'm sorry pma i was lookin at the wrong question.so i apologize

- anonymous

how are you getting \[(a+1)(a+2)\] for LCD @trwatkins1 ?

- triciaal

when one is a factor of the other you do not include

- triciaal

|dw:1439415089426:dw|

- triciaal

oops so sorry I forgot the square when I factored

- triciaal

sorry yes the LCD = (a +1)^2*(a+2)

- Nnesha

don't factor out the GCF
*find the sum* try to submit your answer

- triciaal

|dw:1439415305928:dw|

- triciaal

|dw:1439415500982:dw|

- anonymous

this is a horrible way of learning to factor

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