Question

simplify 2/x^2+8x+15 +1/x^2+11x+30

Answer 1

Collect like terms...

Answer 2

Soccerbabe, you need to make your question a little more clear using brackets.
Is this what it's suppose to look like?\[\large \frac{2}{x^2+8x+15}+\frac{1}{x^2+11x+30}\]

Answer 3

yes that is what it was supposed to look like

Answer 4

Hmm so it looks like we need to factor the bottom polynomials before we can combine these fractions.

Do you know how to break \(x^2+8x+15\) into a pair of binomials? :)

Answer 5

Ah, okay, now you've got to make a common denominator. I would suggest factoring the denominators to make that simpler.

Answer 6

yes it would be \[(x+5)(x+3)\]

Answer 7

And the other one?

Answer 8

Good good c: so we have,\[\large \frac{2}{(x+5)(x+3)}+\frac{1}{x^2+11x+30}\]Can you do it with the other one as well? :)

Answer 9

the other one would be \[(x+5)(x+6)\]?

Answer 10

All right! Someone who doesn't need factoring explained! Let's keep her :-)

Answer 11

lolol ikr :D

Answer 12

Ok good,
\[\large \frac{2}{(x+5)(x+3)}+\frac{1}{(x+5)(x+6)}\]

Answer 13

haha thanks !:)

Answer 14

You're in good hands with @zepdrix, I've got to go.

Answer 15

thank you @whpalmer4 I appreciate it:)

Answer 16

If you were trying to combine these two fractions,\[\large \frac{1}{3\cdot5}+\frac{1}{3\cdot6}\]See how they both have a \(3\)?
We need a 6 in the base of the first fraction, and a 5 in the base of the second.
So we would multiply the first by 5/5, and the second by 6/6.\[\large \frac{6}{6}\frac{1}{3\cdot5}+\frac{1}{3\cdot6}\frac{5}{5} \qquad = \qquad \frac{6}{3\cdot5\cdot6}+\frac{5}{3\cdot5\cdot6}\qquad=\qquad \frac{11}{3\cdot5\cdot6}\]Maybe it's a silly example, but this is the idea, we need to do something like this before we can starting adding things on top.

Answer 17

See in our problem how they both have a factor of \(x+5\) on the bottom?

Answer 18

actually it was a good example and yes I do:)

Answer 19

So that's our "\(3\)", it's the value we don't need to worry about.
But we need the first fraction to have an \(x+6\) in it's base,
and the second to have an \(x+3\) in it's base,

so we'll have have to do some multiplication to get these similar bases :)

Answer 20

so for the x+6 would I need to multiply it by 3 to get it to be 3x+18?

Answer 21

I shouldn't have said 3, I was just trying to make a connection to the example i gave XD lol

Answer 22

\[\large \color{royalblue}{\frac{x+6}{x+6}}\cdot\frac{2}{(x+5)(x+3)}+\frac{1}{(x+5)(x+6)}\cdot \color{royalblue}{\frac{x+3}{x+3}}\]

See how I'm multiplying the top and bottom by the missing factors?

Answer 23

yes.

Answer 24

\[\large \frac{2(x+6)}{(x+6)(x+5)(x+3)}+\frac{1(x+3)}{(x+3)(x+5)(x+6)}\]If this step is confusing, you can let me know.
I put the brackets around them to show that we don't want to expand these terms out. We just want to write them as multiplication.

Answer 25

So then the top of the equation would look like\[3x+15?\]

Answer 26

after you simplify it anyways:)

Answer 27

yes very good :) it looks like we have a tiny bit more simplification we can do.

Answer 28

See how each term on top is divisible by 3?
Try factoring that 3 out, and you might notice a similar term on top that you also have on the bottom! :O

Answer 29

x+5

Answer 30

yah :O interesting!

Answer 31

\[\large \frac{3\color{royalblue}{(x+5)}}{\color{royalblue}{(x+5)}(x+6)(x+3)}\]

Answer 32

So maybe we can do a nice cancellation, yes? :)

Answer 33

yes, yes we can! :)

Answer 34

bahhh i gotta go! D: I think you've got this pretty much figured out though yes? c:

Answer 35

yes I do:) thank you very much for your help! :)

Answer 36

yay team \c:/