Question

Which expression is a simplification of ((x^2+7x+10)/(x^2+2x-3))/(x^2+4x-5)/(x^2-x-12))?
F.(x+4)/X+4 G.x+5/(x+3)(x-4) H. x+2 /x+4
I.(x+2)(x-4)/(x-1)(x-1)

Answer 1

factor each polynomial
invert and multiply, the rational function in the denominator

Answer 2

I forgot how to do eveyrthing.... and what a lot fo those words means sadly

Answer 3

ok

Answer 4

summer hehe

Answer 5

one big one is to take the fraction on the bottom and flip it over to multiply times the first fraction\[\frac{x ^{2}+7x+10}{x ^{2}+2x-3} \frac{x ^{2}-x-12}{x ^{2}+4x-5}\]

Answer 6

now if you factor each quadratic (ax^2+bx+c), then you will be in a good position to cancel things

Answer 7

To factor... take one of the quadratics and look at the last number\[x ^{2}+7x+10\]
now find two numbers that when multiplied equal that number at the end... 10

Answer 8

1X10, 2X5

Answer 9

do i do it for both sides?

Answer 10

if we add those in (both positive or both negative) we get 1+10=11 or 2+5+7

you will need to do it for each quadratic in the problem

Answer 11

I find that 2X5 equals 10, which is the end number. While 2+5 is 7, which is the middle number. So my quadratic factors into (x+2)(x+5).

Answer 12

so would it be? \[(x+2)(x+5)divx ^{2}+2-3 * (X+3)(x-4)divX ^{2}+4X-5\]

Answer 13

now I have\[\frac{(x+2)(x+5)}{x ^{2}+2x-3} \frac{x ^{2}-x-12}{x ^{2}+4x-5}\]yeah, you got the numerator of the second one too! :)

Answer 14

now get the denominators, and start cancelling out the ones in common between the top and the bottom

Answer 15

to make the fractions nice in the editor use frac{numerator}{denominator}

Answer 16

It looks like you're trying to make it prettier, so I thought I would give a suggestion.
Have you factored the denominators?

Answer 17

SORRY I WAS in the bathrookm thanks for your patience!!

Answer 18

no problem... I do that too! :)

Answer 19

okay i think I factored the denominators

Answer 20

what did you get... you can show me the denominators, or what you get after cancelling common factors.

Answer 21

I got \[(x+2)(x+5)\div(x-1)(x+3) * (x+3)(x-4)\div(x-1)(x+5)\]

Answer 22

perfect!

Answer 23

now cancel the ones that appear both above and below... (x+5) and (x+3), the answer will be what is left behind.

Answer 24

so: (x+2)(x-4)/(x-1)(x-1) is what i got

Answer 25

perfect!

Answer 26

yay!!
thanks you!!

Answer 27

You're welcome. Now I have to make dinner for the family.... see ya around.

Answer 28

oh! thank you so much so much responsibility but you still help em i really appreaciate it:)

Answer 29

no problem... click best response button to give me a point ;)

Answer 30

woof!