Question

What is the product in simplest form,state any restrictions.
x^2+5x+6/x+4 * x^2+x-12/x^2+x-2
Help please! I'm drowning in a sea of confusion

Answer 1

This?

\(\Large \frac{x^2+5x+6}{x+4} ~*~ \frac{x^2+x-12}{x^2+x-2} \)

Answer 2

Yes ! exactly that

Answer 3

Alright, first thing you'll need to do is factor, can you do that?

Answer 4

Which part do i factor first?

Answer 5

factor \(x^2+5x+6 \), what do you get?

Answer 6

(x+2) (x+4) I'm not sure how to factor the 5x.
I know you need to add to get both 5x and 6

Answer 7

(x+2) (x+3)

Answer 8

There ya go! :)

Now, factor \(x^2+x-12 \)

Answer 9

(x+6) (x-2) that gives the -12 but not the x in the middle, i just need another moment

Answer 10

Take your time (:

Answer 11

(x-4) (x+3)

Answer 12

switch the positive and negative sign though so (x+4) (x-3)

Answer 13

Close enough :P

That gives us \(x^2-x-12 \) so just change the signs and you're good

I'll do the last one for ya if you don't mind :P

After factoring it all, we get this:

\(\Large \frac{(x+2)(x+3)}{(x+4)}*\frac{(x+4)(x-3)}{(x+2)(x-1) } \)

Now we need to cancel things; notice anything that simplies?

Answer 14

Could you cross out the (x+2) ?

Answer 15

Yes we could!

\(\Large \frac{\cancel{{(x+2)}}(x+3)}{(x+4)}*\frac{(x+4)(x-3)}{\cancel{(x+2)}(x-1) } \)

Anything else?

Answer 16

those damn (x+4)'s xD

Answer 17

Yup

\(\Large \frac{\cancel{{(x+2)}}(x+3)}{\cancel{(x+4)} }*\frac{ \cancel{(x+4)} (x-3)}{\cancel{(x+2)}(x-1) } \)

Which simplies to just:

\(\Large \frac{(x+3)(x-3)}{x-1} \)

Now just gotta find any restrictions (:

Answer 18

3, -3 and 1? or do you not include the factored expression on top

Answer 19

Restrictions are usually just in the denominator, so \(\ x \neq 1 \)

Answer 20

Alright, thank for taking so long to explain this too me CX
I dig your hair too!
Enjoy the rest of the day

Answer 21

Haha, no problem, and thanks, you too :D