in simple form: x^2+7x+12/x-5 times 2x-10/x+3

- anonymous

in simple form: x^2+7x+12/x-5 times 2x-10/x+3

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

the best solution is to factor the "sums" so that you can see the parts that multiplied together to get them.
then you can cancel out like "terms" top to bottom

- amistre64

do you know how to "factor" the sum parts?

- anonymous

No not really I just started this stuff I don't understand it much at all

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

I can try to teach you how to factor if you want.... or do you simply want the answer to the problem :)

- anonymous

Teach me step by step :)

- amistre64

ok..the basic rest in this simple fact...
3(8) = 24
its not the numbers here that are important, but the way we are going to work them that counts.... but Im starting simple.
You agree that 3(8) = 24?

- anonymous

Yes I do agree

- amistre64

good, then this should be the same:
3(6+2) = 24 ... do we agree?

- anonymous

Yes

- amistre64

then the "rule" that allows us to know this is called the "distibutive property". We are going to "distibute" that (3) thru the (6+2) to get our answer like this:
3(6+2) = 24
3(6) + 3(2) = 24
18 + 6 = 24
24 = 24 ..... does that make sense?

- amistre64

distribute is the proper spelling :)

- anonymous

Yes that makes sense :) it's just I don't understand the problem I put down on here

- amistre64

i know...but we have to establish the "rules" for solving this problem since you say that you are unfamiliar with them.... ok?
to solve your problem we need to do the "opposite" if distributing...I simply call it unDistributing.or factoring

- anonymous

Ok I see :)

- amistre64

2x-10 can be unDistributed.... we need to find a numbe that both 2x and 10 have in common so we can pull it out of our way...that number is called the "greatest common factor"...adn simply means the biggest number that they have in common when they are in factored form..
1(2) are the factors of "2"
1(10)
2(5).......are all the factors of 10. now what is the biggest number that they have in common now that they are in factored form?

- anonymous

5?

- amistre64

does "2" have 5 as a factor? I cant see it :)
the biggest number that these have in common would be a "2", that is what they have in common right?

- anonymous

Oh I see now :)

- amistre64

then lets "unDistribute" that 2 and see what we get:
2(x+5) looks about right. we can double check ourselves by "distributing" it together again:
2(x+5)
2(x) + 2(5)
2x + 10 ...yep that works.... you see it?

- anonymous

Yes

- amistre64

good, then we can try our hands at that "quadratic" expression... quadratic is just some fancy latin name meaning "highest exponent is a '2'...

- anonymous

ok good :)

- amistre64

And I see a typo I made above..... 2(x-5) is what we wanted from that.......
x^2+7x+12 is tricker but it has the same "basic" concept as before...there are 2 "numbers" that multiply together to get this expression. but they are in the form of:
(1+2)(6+2) = 24 which is more complicated...

- amistre64

that +12 hanging out on the end...we need to know all the ways we can factor that out.
1(12) = 12
2(6) = 12
3(4) = 12 ....these are our possible options to work with, do you agree?

- anonymous

yes I do agree completely :)

- amistre64

yay!!
you see that middle term? that (+7)? it wants 2 numbers that are added together to make it a +7.
Lets try to see if we can use our "pool" of numbers to get us some answers:
1+12 = 13 ....not 7
2+6 = 8 ......still not 7
3 + 4 = 7 ...... we found a winner!!

- amistre64

now we need to set it up like this:
(x+3)(x+4)
double check ourselves:
(x+3)(x+4)
(x+3)(x) + (x+3)(4)
xx + 3x + 4x +3(4)
x^2 + 7x + 12 .... it fits...

- anonymous

ok

- amistre64

like I said...more complicated, but still the same basic rules....
so we have this now:
(x+3)(x+4) (2)(x-5)
--------- ------
(x-5) (x+3)
any like "terms" cancel each other from top to bottom. That means each (.....) that is the same on top and on bottom can be crossed out.
(x+3)(x+4) (2)(x-5)
--------- ------
(x-5) (x+3) what can we cross out?

- anonymous

(x+3) and (x+5)

- amistre64

thats right.... adjusting for typos that is ;)
so lets get rid of those "values" and what we have left is:
(x+4)(2) which is a perfectly good setup, but some tests may want you to write it out in its "distributed form:
2(x) + 2(4)
2x + 8 is another way of writing it

- anonymous

Ok I see I think I have the answer to the problem now is it 2(x+4)? that's what my worksheet said. I just had to show my work to get the write answer.

- anonymous

right I mean*

- amistre64

2(x+4) is a fine answer..... and is accepted at all the finer establishments thru town :)

- anonymous

Yay! I got it correct! :)

- amistre64

yay!!

- anonymous

I need help again on this problem:
x^2-3x-10/x+7 times 3x+12/6x-30

Looking for something else?

Not the answer you are looking for? Search for more explanations.