Question

3v^2-20x-32
_______________
2x^2-x-15
__________________
9x^2+24x+16
________________
2x^2+7x+5

Answer 1

does anybody out there understand this?

Answer 2

Is that all one gigantic fraction?

\[ \huge \dfrac{\left( \frac{3v^2-20x-32}{2x^2-x-15}\right) }{\left( \frac{9x^2+24x+16}{2x^2+7x+5}\right) } \]

Answer 3

yes

Answer 4

Goodness gracious, you got a lot of factoring to do. :-p FIRST though, "invert & Multiply.

\[ \huge \left( \frac{3v^2-20x-32}{2x^2-x-15}\right) \cdot \left( \frac{2x^2+7x+5}{9x^2+24x+16} \right) \]

Answer 5

NOW factor each expression individually. Can you factor

\[ \huge 3v^2-20x -32\]
for instance?

Answer 6

I am a bit confused by it all

Answer 7

yes, and thats supposed to be an x

Answer 8

This can certainly be overwhelming if you're not familiar with four main things:

1. Simplifying simple fractions. (with numbers, no variables)
2. Simplifying simple fractions (with variables)
3. Factoring quadratic equations
4. Simplifying complex fractions (with numbers, no variables)

Answer 9

yes i am in over my head but trying to figur it out

Answer 10

ok so if I get it all factored, whats next?

Answer 11

You go cancel-crazy. If you have the same term in the numerator and denominator, cancel it.

But first, let's talk about factoring quadratic equations...

3v^2-20x-32

Can you factor that? Just, focus on that only right now. :)

Answer 12

(3x+8 )(x-4 )??

Answer 13

The x terms don't work out.

8x -12x = -4x

Answer 14

oh (3x+4)(x-8)

Answer 15

Yes, you got it. :)

3x*x = 3x^2

4x -24x = -20x

4*(-8) = -32

Answer 16

Ok, now, can you factor

2x^2-x-15

Answer 17

so is the key to this whole factoring thing just trial and error? or is there some magic concept that can be used?

Answer 18

(2x+ )(x- )
I honestly dont know what I should use here

Answer 19

It's just practice, really. You start to spot patterns. That's a good start.

Next step = What are the factors of 15 ?

Answer 20

5&3 or 15&1

Answer 21

Yup. So the key is experimenting with those. The answer is somewhere in there. Just takes some playing around with. :)

Answer 22

(2x-1)(x+15)?

Answer 23

The x terms don't work out

-1x + 30x = 29x

Answer 24

yeah this is where i get confused a bit... the -1 makes the last part right but not the middle number

Answer 25

Hint: 1 and 15 won't do it. No matter how you try, it won't work.

Answer 26

ok...so (2x+3)(x-5)

Answer 27

ahh (2x+5)(x-3)

Answer 28

hey thanks for your help and your patience! i just dont get this stuff... my brain doesnt work this way... i will probably never understand it

Answer 29

Bingo.

Answer 30

ok so the 2nd part i just do the same thing... but then what? do i have to invert something there?

Answer 31

So the key is to factor all four quadratics. I already "flipped and multiplied".

\[\huge \left( \frac{3v^2-20x-32}{2x^2-x-15}\right) \cdot \left( \frac{2x^2+7x+5}{9x^2+24x+16} \right)\]

Then when you have everything factored, cancel whatever you can.
\[\huge \frac{( )( )}{()()}\cdot \frac{()()}{()()}\]

Answer 32

ok thanks a bunch!

Answer 33

You're welcome! :) Have a good evening.

Answer 34

you too :)