Question

Algebra 2 simplifying rational expressions

Answer 1

@IMStuck

Answer 2

for the first one I got 2(x^2+5x+3)/(x+3)(x+5)?

Answer 3

Yep or you could leave it with the 2 factored into it and the denominator FOILed out. But yes, that's correct.

Answer 4

ok I didn't do number 2 yet

Answer 5

Are you attempting #2? It's really nice and simple, this one is.

Answer 6

yes hold on haha

Answer 7

um x+4 cancels out and so does x +3

Answer 8

Yep, so what are you left with then?

Answer 9

(x+2)(x-4) on the bottom but does a one go on the top because the numerators both got cancelled out. if so then its 1/x^2-8?

Answer 10

Yes, the 1 goes on top. Just because you cancelled out the factors, you can't disregard the fact that it was a fraction to begin with (even though we would probably all prefer to pretend fractions didn't exist!) The 1 is the placeholder for the factors you cancelled. Very good!!!!

Answer 11

What about #3? What do you think you need to do with that?

Answer 12

factor

Answer 13

Right. Factor both the denominators and you will probably see one or two factors in common between them. Let me know what you get when you have them both factored.

Answer 14

so 2/(x-3)(x+3) - 3x/(x-3)(x-2)

cancel out x-3

then you are left with 2/x+3 - 3x/x-2

then what?

Answer 15

No back up. You can't cancel anything out. You can only cancel like that when there is an = sign in between the terms; here there is a - sign. Do Not Cancel here.

Answer 16

ohhhh sorry sorry I forgot

Answer 17

The first term has an (x + 3) and a (x - 3) in the denominator; the second term has an (x - 3) and a (x - 2). In order to get a common denominator, you have to multiply a (x - 2) into the top and bottom of the expression:\[\frac{ (x - 2) }{ (x - 2) }\frac{ 2 }{ (x +3)(x-3) }\]When you do that you get a denominator of (x - 2)(x + 3)(x - 3) with 2(x - 2) in the numerator. Get that?

Answer 18

yes

Answer 19

Now the second term needs an (x + 3) to make it common with the other denominator so we will multiply that by the top and bottom:\[\frac{ (x+3) }{ (x+3) }\frac{ 3x }{ (x-3)(x-2) }\]Now that denominator is (x + 3)(x - 3)(x - 2) just like the other one. Now here's what you have all over one common denom.\[\frac{ 2(x-2)-[3x(x+3)] }{ (x-2)(x+3)(x-3) }\]

Answer 20

ok then what?

Answer 21

Doing the math you get\[\frac{ 2x-4-(3x ^{2}+9) }{ (x-2)(x-3)(x+3) }\]Now of course you have that pesty minus sign there so that means that the positive sign in front of the 9 will be changed to a negative sign:\[\frac{ 2x-4-3x ^{2}-9 }{ (x-2)(x-3)(x+3) }\]The numerator simplifies to \[-3x ^{2}+2x-13\]So your expression is\[-\frac{ 3x ^{2}+2x-13 }{ (x-2)(x-3)(x+3) }\]That's it; believe it or not that is simplified!!!!

Answer 22

Sweet thank you :) ok #4 :P

Answer 23

Number 4 is not nearly as ugly or involved as that one was!

Answer 24

You know that when you see a division sign here that you have to flip the second fraction and change the sign to a multiplication sign, right?

Answer 25

yes mam

Answer 26

Then factor the denominator in the first expression and then factor the x^2 - 16 in the other one. This is what you get then:\[\frac{ x+4 }{ (x-3)(x-2) }\times \frac{ x+3 }{ (x+4)(x-4) }\]Now you can cross cancel between the (x + 4)'s. That's all that can cancel out here.

Answer 27

so x+3/(x-3)(x-2)(x-4)?

Answer 28

Perfect!!!!!!!

Answer 29

sweet ok #5 is just weird

Answer 30

there are more steps with rational expressions than just integer?

Answer 31

Well they have in common the fact that both can be put into fraction form and then divided out. There are definitely more steps because you have to worry about with rational expressions, but the major difference is that when you are dividing rational expressions, you flip the fraction and change the sign. That's the one they want you to notice. That's the most important thing.

Answer 32

I meant that you have to worry about number and variables with rational expressions (I didn't quite finish my sentence there!).

Answer 33

Get those now? Task 4 all done?

Answer 34

yess and now im mad... task 5 is the hardest and longest one...

Answer 35

Post it for me...then I have to take a break to start dinner for my family. Post it and I will look at it and get my act together and come back to you. Is that ok?

Answer 36

well I kinda have to write a letter regarding whether or not algebra 2 honors is necessary... ill do this one on my own... Thanks for everything! You helped me understand so much today thanks!

Answer 37

OMG! A letter, huh? Algebra 2 is definitely one of the most challenging math topics there is...next to calculus. But I love Algebra and Algebra 2! And you're welcome! It was fun! You're a pretty smart kid!!!!! Keep up the good work!