anonymous
  • anonymous
Adding rational expressions. I keep getting these wrong. x/x^2+6x+9 + 7/x^2+8x+15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Nnesha
  • Nnesha
\[\frac{ x }{ x^2+6x+7 } +\frac{ 7 }{ x^2+8x+15 }\] like this right ?
anonymous
  • anonymous
that was dauntingly fast !
anonymous
  • anonymous
x^2+6x+9 on the first one

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anonymous
  • anonymous
5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again.
anonymous
  • 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
  • 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
  • Nnesha
okay gimme a sec :)
Nnesha
  • Nnesha
so so what are the factors of x^2+6x+9 ??
Nnesha
  • Nnesha
so so what are the factors of x^2+8x+15??
anonymous
  • 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
  • Nnesha
hmm okay good :)
Nnesha
  • Nnesha
do you know how to find factors quadratic equations ?
Nnesha
  • Nnesha
* of
anonymous
  • anonymous
No, i don't
Nnesha
  • 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
  • anonymous
I solved it with factoring and it was correct.
Nnesha
  • 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
  • anonymous
I redid it and got something different
anonymous
  • anonymous
I have another question now. Can we try this one?
Nnesha
  • Nnesha
alright! :) post
Nnesha
  • Nnesha
well answer for that one is incorrect we should try it again!
anonymous
  • anonymous
2y/y^2 -2y+1 + 8/y^2+1y-2
anonymous
  • anonymous
I got the previous question correct once i redid it.
Nnesha
  • Nnesha
ohh okay.
Nnesha
  • Nnesha
have you tried to solve the 2nd one ?
anonymous
  • anonymous
yes. i'm getting 2(y^2+6y-4)/(y-1)^2(y+2)
anonymous
  • anonymous
but I know it'll be wrong if I put it as my answer.
anonymous
  • anonymous
is it not factored?
Nnesha
  • 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
  • Nnesha
it's factorable
anonymous
  • anonymous
i think i need to put in the unfactored answer
anonymous
  • anonymous
because anytime i put it in factored, it's wrong
Nnesha
  • Nnesha
1st) factor quadratic equation 2nd) find common denominator
Nnesha
  • 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
  • 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
  • anonymous
(y-1)^2(y+2)
anonymous
  • anonymous
i have no clue what im doing
Nnesha
  • 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
  • anonymous
it says it was incorrect.
Nnesha
  • 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
  • anonymous
4x/x^2+2x+1 + 3/x^2+3x+2 is the new question
anonymous
  • anonymous
4x^2 +11x+3/(x+1)^2(x+2 is the answer im getting
anonymous
  • anonymous
and i did not. i submitted my answer before you said to take out the gcf so it was incorrect
anonymous
  • anonymous
1 is the gcf, right?
anonymous
  • anonymous
im confused
anonymous
  • anonymous
so do I take 2 out of my final answer?
Nnesha
  • 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
  • anonymous
4x^2+11x+3/(x+1)^2x
anonymous
  • anonymous
is that then?
Nnesha
  • 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
  • anonymous
i am completely lost
Nnesha
  • Nnesha
please type the number front of the new question
Nnesha
  • Nnesha
like 2nd ) question 3rd) question
anonymous
  • anonymous
i have no clue what question im on
anonymous
  • anonymous
im giving up.
anonymous
  • anonymous
never give up, take a breath get a drink
anonymous
  • anonymous
thanks for your help, though. i appreciate it a lot.
Nnesha
  • Nnesha
relax! you are doing a great job!
Nnesha
  • Nnesha
alright i didn't give u a GREAT explanation so yeah i'll tag other users
Nnesha
  • Nnesha
let's see..
anonymous
  • anonymous
@triciaal 's great
Nnesha
  • Nnesha
@Hero @triciaal @nincompoop
anonymous
  • anonymous
is the problem factoring ?
anonymous
  • 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
  • Nnesha
i would like to see the screenshot of *multiple choice or answer box what is the statement ??
anonymous
  • anonymous
okay. how can i enter a photo?
Nnesha
  • Nnesha
Press PrntScr to take a screenshot and do you see to blue *attach file* button to attach file ?
anonymous
  • anonymous
I don't see an attach file button
anonymous
  • anonymous
1 Attachment
Nnesha
  • Nnesha
http://prntscr.com/83xios
Nnesha
  • Nnesha
answer is not correct
Nnesha
  • Nnesha
show your work so i can loook over and find out some mistakes that would be great
anonymous
  • 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
  • triciaal
do you still need help?
anonymous
  • anonymous
I got locked out. My account is no longer active.
anonymous
  • 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
  • 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
  • Nnesha
ik numerator isn't correct so that's mean answer is not correct
anonymous
  • anonymous
well the numerator is correct (again i might be wrong ) just not factored
triciaal
  • triciaal
|dw:1439414757668:dw|
Nnesha
  • Nnesha
alright let's see \[ \frac{ 6a }{ a^2+2a+1 } +\frac{ 7 }{ a^2+3a+2 }\]
Nnesha
  • Nnesha
ohhhhhh i see *facepalm* it's 2 not 7
triciaal
  • triciaal
so your denominator is wrong
Nnesha
  • Nnesha
was looking at the wrong question.
anonymous
  • anonymous
im pretty sure LCD is (a+2)(a+1)^2
Nnesha
  • Nnesha
ye answer is correct i'm sorry pma i was lookin at the wrong question.so i apologize
anonymous
  • anonymous
how are you getting \[(a+1)(a+2)\] for LCD @trwatkins1 ?
triciaal
  • triciaal
when one is a factor of the other you do not include
triciaal
  • triciaal
|dw:1439415089426:dw|
triciaal
  • triciaal
oops so sorry I forgot the square when I factored
triciaal
  • triciaal
sorry yes the LCD = (a +1)^2*(a+2)
Nnesha
  • Nnesha
don't factor out the GCF *find the sum* try to submit your answer
triciaal
  • triciaal
|dw:1439415305928:dw|
triciaal
  • triciaal
|dw:1439415500982:dw|
anonymous
  • anonymous
this is a horrible way of learning to factor

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