Question

x^2+13x+40 over x-7 divided by x+8 over x^2-49

Answer 1

\[\Large \frac{ \frac{ x^2+13x+40 }{ x-7 } }{ \frac{ x+8 }{ x^2-49 } }\]

Answer 2

since you have some thing like that,
you would have to know that,
\[\Large \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } }~=~\frac{ a }{ b }\times \frac{ d }{ c }\]

Answer 3

using that, what do you get for the 1st step??

Answer 4

Here
http://assets.openstudy.com/updates/attachments/51aeeee1e4b06ee3ee2668a8-pyromancer-1370421337529-unitreviewqsonlyoddshowwork.jpg

Answer 5

#13

Answer 6

Are you there?

Answer 7

oh....it is the same as what i have written
\[\Large \frac{ x^2+13x+40 }{ x-7 }\div \frac{ x+8 }{ x^2-49 }\]

Answer 8

Yes

Answer 9

So whats the 1st step

Answer 10

okay, so with that said
you would have to know that
\[\frac{ a }{ b }\div \frac{ c }{ d }=\frac{ a }{ b }\times \frac{ d }{ c }\]

Answer 11

ok
so
\[same thing here \div \frac{ x+8 }{ x ^{2}-49 }\]

Answer 12

oops i meant to flip it lol I'm such a dunce

Answer 13

you would have to introduce the multiplication sign as you flip it..

Answer 14

oh yea

Answer 15

so then what?

Answer 16

factorise x^2+13x+40

Answer 17

what do you get??

Answer 18

http://wolframalpha.com

Answer 19

what do you get?

Answer 20

(x+5)(x+8)

Answer 21

that is right.......are you able to identify x^2-49 as difference of two squares??

Answer 22

what do you mean?

Answer 23

difference of two squares..
\[a^2-b^2=(a+b)(a-b)\]

Answer 24

do you know that?

Answer 25

ok now i do..which #'s are a and b?

Answer 26

let me use it in an example:
\[y^2-4=(y+2)(y-2)\]

Answer 27

do you get it a bit now??

Answer 28

oh so what i just did (x+5) and (x+8)

Answer 29

i do not get what you are saying..

Answer 30

factorise x^2+13x+40 equals what i just said

Answer 31

Could you do this part?

Answer 32

yes....
so if you apply the difference of two square to x^2-49 what do you get?

Answer 33

do you get it or you are a bit confused?

Answer 34

(x+5)-(x+8)?

Answer 35

\[\Large \frac{ (x+5)(x+8) }{ x-7 } \times \frac{ x^2-49 }{ x+8 }\]

Answer 36

ok

Answer 37

you would have to factorise x^2-49
and to do that, difference of two squares is used
because both the x^2 and 49 are perfect and squares and are subtracting..

Answer 38

do you understand it now?

Answer 39

ok yes...then what would you get?

Answer 40

i gave an example above try to see whether you can figure it out..

Answer 41

example: given
\[y^2-4\]
this could be rewritten as:
\[y^2-2^2\]
and using the fact i gave for factorising difference of two squares,
\[(y+2)(y-2)\]

Answer 42

ok now i unders

Answer 43

understand*

Answer 44

okay..
so what do you get when you do the same for x^2-49??

Answer 45

\[x ^{2}-7^{2}=(x+7)(x-7)\]

Answer 46

?hmmm?Was I right?

Answer 47

yes!....

Answer 48

so plug in the factorised versions of the x^2+13x+40 and x^2-49 into the question and what do you get?

Answer 49

ok now can you right out whole problem so i can put it down on paper now that i understand (and maybe a medal too)?

Answer 50

Oh ok

Answer 51

i don't get what you are saying...

Answer 52

x^2+13x+40/x-7 times (x+7)(x-7)/x+8?

Answer 53

you could just write everything at once
\[\Large \frac{ (x+5)(x+8) }{ x-7 }\times \frac{(x+7)(x-7) }{ x+8 }\]

Answer 54

yay cool

Answer 55

ok my next question will be the last one like this. :D

Answer 56

notice some cancel out
what do yuo get after cancelling out?

Answer 57

**you

Answer 58

Oh...such as?

Answer 59

(which 1s cancel out there are no common denominators..oh the 7's?

Answer 60

so x^2 over x+8?