Question

Divide. Can someone please help?

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Answer 1

@jim_thompson5910 do you know how to do this ?

Answer 2

The first step is to flip the second fraction and multiply

so that gives you

\[\Large \frac{x^2+2x+1}{x-2} \bullet \frac{x^2-4}{x^2-1}\]

what's next?

Answer 3

ummm i really have no idea...

Answer 4

the next step is to factor each expression

so x^2+2x+1 factors to ??

Answer 5

i am not understanding :(

Answer 6

are you able to factor x^2+2x+1

Answer 7

do you set it equal to zero?

Answer 8

have a look at this page

http://www.purplemath.com/modules/factquad.htm

Answer 9

it gives an example

Answer 10

would it be (x + 1)(x + 1)?

Answer 11

good

Answer 12

factor x^2-4 to get ???

Answer 13

it might help to think of x^2-4 as x^2+0x-4

Answer 14

(x-2)(x+2).?

Answer 15

?????????

Answer 16

good, then x^2 - 1 factors to ???

Answer 17

(x - 1)(x + 1) ???

Answer 18

So

\[\Large \frac{x^2+2x+1}{x-2} \bullet \frac{x^2-4}{x^2-1}\]

turns into

\[\Large \frac{(x+1)(x+1)}{x-2} \bullet \frac{(x-2)(x+2)}{(x-1)(x+1)}\]

Answer 19

is that the answer ?

Answer 20

then you do these cancellations

\[\Large \frac{(x+1)(x+1)}{x-2} \bullet \frac{(x-2)(x+2)}{(x-1)(x+1)}\]

\[\Large \frac{\cancel{(x+1)}(x+1)}{x-2} \bullet \frac{(x-2)(x+2)}{(x-1)\cancel{(x+1)}}\]

\[\Large \frac{x+1}{x-2} \bullet \frac{(x-2)(x+2)}{x-1}\]

\[\Large \frac{x+1}{\cancel{x-2}} \bullet \frac{\cancel{(x-2)}(x+2)}{x-1}\]

\[\Large \frac{x+1}{1} \bullet \frac{x+2}{x-1}\]

\[\Large \frac{(x+1)(x+2)}{x-1}\]

Answer 21

The cancellations work because dividing any expression by itself is 1 (example: 5/5 = 1)

Answer 22

Then 1*x = x, so the 1 effectively goes away

Answer 23

ok

Answer 24

can you help with this one ? simplify the rational expression. state any excluded values. x^2-x-6/x+2

Answer 25

factor the numerator to get ???

Answer 26

ummm (x+2) (x+2) idk this one

Answer 27

x^2-x-6 is the same as x^2-1x-6

so you have to find two numbers that

multiply to -6
AND
add to -1

Answer 28

@jim_thompson5910 It amazes me how well you teach. Sorry to interrupt.

Answer 29

Thank you for the compliment. It's greatly appreciated.

Answer 30

A.x - 3, where x 3
B. x-2
C.1/x- 2 , where x - 2
D.x - 3, where x -2

these are the answers i can choice from and and i am think it is B? is that correct or is it C?

Answer 31

wait i think i did my math wrong....

Answer 32

what did you get when you factored x^2-1x-6

Answer 33

it would be (x-3)(x-3) right?

Answer 34

-3 times -3 = 9

we want the two numbers to multiply to -6

Answer 35

(x-3)(-2)?

Answer 36

I think you meant (x-3)(x-2)

Answer 37

-3 times -2 = 6

again, we want -6

Answer 38

i am confused?

Answer 39

here are all of the ways to multiply to -6


-6 = -1*6
-6 = -2*3
-6 = -3*2
-6 = -6*1


which pair of factors adds to -1 ?

Answer 40

-3+2= -1

Answer 41

that -3 and +2 are then used to form the factorization of x^2-1x-6

Answer 42

x^2-1x-6 factors to (x-3)(x+2)

Answer 43

\[\Large \frac{x^2-x-6}{x+2}\]

turns into

\[\Large \frac{(x-3)(x+2)}{x+2}\]

I'm sure you see what to do from here

Answer 44

(x-3)
but out of the answers to pick what one would it be
A.x - 3, where x neq 3
B. x-2
C.1/x- 2 , where x neq - 2
D.x - 3, where x neq -2

Answer 45

keep in mind that you cannot divide by zero

Answer 46

so you have to make sure the denominator isn't zero

the denominator is zero when x = -2 since

x+2 = 0
x = -2

so we must add the restriction that \(\Large x \neq -2\) to avoid division by zero.

Answer 47

x can be 3 since that doesn't make the denominator zero

Answer 48

so it would be D ?

Answer 49

yes

Answer 50

ok thank you!

Answer 51

np