anonymous
  • anonymous
in simple form: x^2+7x+12/x-5 times 2x-10/x+3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • 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
  • amistre64
do you know how to "factor" the sum parts?
anonymous
  • anonymous
No not really I just started this stuff I don't understand it much at all

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amistre64
  • amistre64
I can try to teach you how to factor if you want.... or do you simply want the answer to the problem :)
anonymous
  • anonymous
Teach me step by step :)
amistre64
  • 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
  • anonymous
Yes I do agree
amistre64
  • amistre64
good, then this should be the same: 3(6+2) = 24 ... do we agree?
anonymous
  • anonymous
Yes
amistre64
  • 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
  • amistre64
distribute is the proper spelling :)
anonymous
  • anonymous
Yes that makes sense :) it's just I don't understand the problem I put down on here
amistre64
  • 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
  • anonymous
Ok I see :)
amistre64
  • 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
  • anonymous
5?
amistre64
  • 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
  • anonymous
Oh I see now :)
amistre64
  • 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
  • anonymous
Yes
amistre64
  • 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
  • anonymous
ok good :)
amistre64
  • 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
  • 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
  • anonymous
yes I do agree completely :)
amistre64
  • 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
  • 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
  • anonymous
ok
amistre64
  • 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
  • anonymous
(x+3) and (x+5)
amistre64
  • 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
  • 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
  • anonymous
right I mean*
amistre64
  • amistre64
2(x+4) is a fine answer..... and is accepted at all the finer establishments thru town :)
anonymous
  • anonymous
Yay! I got it correct! :)
amistre64
  • amistre64
yay!!
anonymous
  • anonymous
I need help again on this problem: x^2-3x-10/x+7 times 3x+12/6x-30

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