Question

will medal and fan! Could somebody show me how to work this out? (question in comments)

Answer 1

@sand-lock53 ...
so recall that when you are dividing two fractions.. you will keep the first one the same and than flip the next fraction * & change the division sign to a multiplication sign
*Keep, flip, change

Answer 2

what about the factoring part?

Answer 3

\[\frac{ x^2-4 }{ x^2-16 }\div \frac{ x^2+8x-20 }{ x-4 }\]

Answer 4

@sand-lock53 yep you also factor too haha and cancel terms that are alike

Answer 5

It's the factoring that confuses me :p

Answer 6

@sand-lock53 hmm... do you have any work that I can look at?

Answer 7

yeah, hold on

Answer 8

so far i have (x-4)(x-4) / (x-4)(x+4)(x-2)(x+10 does this look good?

Answer 9

Hmm.... so
\[\frac{ (x-2)(x+2) }{ (x+4)(x-4)} * \frac{ (x-4) }{ (x-10)(x+2) }\]
you would eventually factor it out to this.. than you can cross cancel factors

Answer 10

*flip it first

Answer 11

the x+2 & the x-4 cancel...

Answer 12

yes, that's what i have (sorry if it was confusing). I thought it was factors with only the same numbers (eg (x+2) & (x+2)), but I guess it's not.

Answer 13

oooppss factored wrong*** sorry haha

Answer 14

and no worries...

Answer 15

\[\frac{ (x-2)(x+2) }{ (x-4)(x+4) }*\frac{ (x-4) }{ (x-2)(x+10) }\]

Answer 16

**** OK so the x-2 & the x-4 cancel..

Answer 17

you get -->
\[\frac{ (x+2) }{ (x+10)(x+4)} \]

Answer 18

as far as the parenthesis go, does order matter?

Answer 19

expand-->
\[\frac{ x+2 }{ x^2+14x+40}\]

and hmmm @sand-lock53 what do you mean?

Answer 20

in one of the previous equations, you switched (x-/+10) from the left side, to the right side. is there always a specific order that they have to go in?

Answer 21

oh.. that was MY factoring mistake sorry to confuse you.
\[x^2+8x-20=(x+10)(x-2)\]

Answer 22

oh, ok. got it

Answer 23

do i have to reduce the fraction? x+2 / (x^2)+14x+40

Answer 24

nope. that is the most simplified that is going to get XD haha

Answer 25

well, that explains why this was so hard. Anyways, thank you so much for your help, I appreciate it! :)

Answer 26

NP always happy to help :D

Answer 27

Here is the problem:

\(\Large \dfrac{x^2-4}{x^2 - 16} \div \dfrac{x^2 + 8x - 20}{x - 4}=\)

Step 1: change the division into a multiplication by writing times instead if divided by and by replacing the second fraction with its reciprocal. In other words, switch from divide to multiply, and flip the second fraction.

\(\Large =\dfrac{\color{red}{x^2-4}}{\color{green}{x^2 - 16}} \times \dfrac{x - 4}{\color{blue}{x^2 + 8x - 20}} \)

Step 2. Factor everything that is factorable. Each factorable polynomial is shown below factored in the same color it was shown above

\(\Large =\dfrac{\color{red}{(x+2)(x-2)}}{\color{green}{(x+4)(x-4)}} \times \dfrac{x - 4}{\color{blue}{(x+10)(x-2)}} \)

Step 3. Now before we multiply, we divide the numerator by the common factors (usually called canceling out common factors)

\(\Large =\dfrac{(x+2)(\cancel{x-2})}{(x+4)(\cancel{x-4})} \times \dfrac{\cancel{x - 4}}{(x+10)(\cancel{x-2})} \)

Step 4. Multiply the fractions

\(\Large =\dfrac{(x+2)}{(x+4)(x+10)} \)

You can leave it factored as shown above.

If you are asked to multiply out the denominator, then you can use FOIL to get

\(\Large =\dfrac{x + 2}{x^2+10x +4x+40} \)

\(\Large =\dfrac{x + 2}{x^2+14x +40} \)

Answer 28

@mathstudent55 so neat :O

Answer 29

Thanks.