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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\frac{ x }{ x^2+6x+7 } +\frac{ 7 }{ x^2+8x+15 }\] like this right ?
that was dauntingly fast !
x^2+6x+9 on the first one

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again.
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
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
okay gimme a sec :)
so so what are the factors of x^2+6x+9 ??
so so what are the factors of x^2+8x+15??
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.
hmm okay good :)
do you know how to find factors quadratic equations ?
* of
No, i don't
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
I solved it with factoring and it was correct.
\(\color{blue}{\text{Originally Posted by}}\) @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
I redid it and got something different
I have another question now. Can we try this one?
alright! :) post
well answer for that one is incorrect we should try it again!
2y/y^2 -2y+1 + 8/y^2+1y-2
I got the previous question correct once i redid it.
ohh okay.
have you tried to solve the 2nd one ?
yes. i'm getting 2(y^2+6y-4)/(y-1)^2(y+2)
but I know it'll be wrong if I put it as my answer.
is it not factored?
\[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\] 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
it's factorable
i think i need to put in the unfactored answer
because anytime i put it in factored, it's wrong
1st) factor quadratic equation 2nd) find common denominator
\(\color{blue}{\text{Originally Posted by}}\) @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)
\[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\] \[\rm \frac{ 2y }{ \color{reD}{(y-1)}(y-1) } +\frac{ 8 }{ (y+2)\color{ReD}{(y-1)} }\] (y+2) (y-1) are factors of y^2+y-2 now what is the common denominator ?
(y-1)^2(y+2)
i have no clue what im doing
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
it says it was incorrect.
\[\rm \frac{ 2y^2+12y-8 }{ (y-1)^2(y+2) } \] =\[\frac{ 2(y^2+6y-4) }{ (y-1)^2(y+2)}\] take out the GCF(greatest common factor
4x/x^2+2x+1 + 3/x^2+3x+2 is the new question
4x^2 +11x+3/(x+1)^2(x+2 is the answer im getting
and i did not. i submitted my answer before you said to take out the gcf so it was incorrect
1 is the gcf, right?
im confused
so do I take 2 out of my final answer?
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
4x^2+11x+3/(x+1)^2x
is that then?
\(\color{blue}{\text{Originally Posted by}}\) @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!
i am completely lost
please type the number front of the new question
like 2nd ) question 3rd) question
i have no clue what question im on
im giving up.
never give up, take a breath get a drink
thanks for your help, though. i appreciate it a lot.
relax! you are doing a great job!
alright i didn't give u a GREAT explanation so yeah i'll tag other users
let's see..
@triciaal 's great
is the problem factoring ?
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.
i would like to see the screenshot of *multiple choice or answer box what is the statement ??
okay. how can i enter a photo?
Press PrntScr to take a screenshot and do you see to blue *attach file* button to attach file ?
I don't see an attach file button
1 Attachment
http://prntscr.com/83xios
answer is not correct
show your work so i can loook over and find out some mistakes that would be great
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.
do you still need help?
I got locked out. My account is no longer active.
besides factoring the nominator i don't see anything wrong (i'm might be doing it wrong too lol) what exactly is wrong @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
ik numerator isn't correct so that's mean answer is not correct
well the numerator is correct (again i might be wrong ) just not factored
|dw:1439414757668:dw|
alright let's see \[ \frac{ 6a }{ a^2+2a+1 } +\frac{ 7 }{ a^2+3a+2 }\]
ohhhhhh i see *facepalm* it's 2 not 7
so your denominator is wrong
was looking at the wrong question.
im pretty sure LCD is (a+2)(a+1)^2
ye answer is correct i'm sorry pma i was lookin at the wrong question.so i apologize
how are you getting \[(a+1)(a+2)\] for LCD @trwatkins1 ?
when one is a factor of the other you do not include
|dw:1439415089426:dw|
oops so sorry I forgot the square when I factored
sorry yes the LCD = (a +1)^2*(a+2)
don't factor out the GCF *find the sum* try to submit your answer
|dw:1439415305928:dw|
|dw:1439415500982:dw|
this is a horrible way of learning to factor

Not the answer you are looking for?

Search for more explanations.

Ask your own question