## chicagochica5 3 years ago Divide.

1. chicagochica5

2. miteshchvm

the x^2 + 2x + 1 is thw square of (x + 1)^2 as per the previous questions you can solve it

3. chicagochica5

The first part would be (x -2) 2x? + 1

Yes!

5. chicagochica5

@theredhead1617 do you know how to solve the problem

6. miteshchvm

$\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }$ $\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\$ \frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\]\] can you solv further?

7. chicagochica5

No @miteshchvm I dont understand how to. I wish to know where to begin but i do not

8. miteshchvm

can you factorise x^2 + 2x + 1?

9. miteshchvm

x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) x^2 - 1 = (x +1)(x - 1) x^2 - 4 = (x-2)(x+2) you get it?

10. chicagochica5

(x + 1) to second power?

11. miteshchvm

yes

12. miteshchvm

$x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1)$ $x^2 - 1 = (x +1)(x - 1)$ $x^2 - 4 = (x-2)(x+2)$ you get it? i wrote this instead of your question in order to simplify,

13. chicagochica5

just a little bit, its hard

14. miteshchvm

now refer to my first comment

15. chicagochica5

on the previous question?

16. miteshchvm

your question can be written as${ \frac{ (x^2 + 2x + 1) }{ x-2 } } \div { \frac{ x^2 - 1 }{ x^2 - 4 } }$ now $\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\$ \] $\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\$ \]

17. miteshchvm

you get it?

18. chicagochica5

im getting a better understand of it

19. chicagochica5

A:

20. chicagochica5

B:

21. chicagochica5

C:

22. chicagochica5

D:

23. miteshchvm

on eliminating values you get $\frac{ (x+1)(x+2) }{ (x-1) }$

24. chicagochica5

thanks for everything @miteshchvm