## monroe17 3 years ago ((x^2+3x+4)-2) simplifies to -2x^2-6x-8

1. monroe17

fog.. when f(x)=x-2 g(x)=x^2+3x+4 =((x^2+3x+4)-2) = -2x^2-6x-8

2. jim_thompson5910

f(x)=x-2 f(g(x))=g(x)-2 f(g(x))=x^2+3x+4-2 f(g(x))=x^2+3x+2

3. jim_thompson5910

You're not multiplying, you're subtracting in this case

4. monroe17

oh.. so I don't distribute the -2?

5. monroe17

so for fof.. would is be.. (x-2)-2=x-4?

6. jim_thompson5910

no, you're not distributing since it's originally x-2 and not -2x

7. jim_thompson5910

yes, fof is x-4

8. 85295james

no the roots is X=-1 or -2

9. monroe17

10. jim_thompson5910

(x-2)^2+3(x-2)+4 x^2-4x+4+3(x-2)+4 x^2-4x+4+3x-6+4 x^2-x+2

11. 85295james

the polynomial discriminant: =1

12. 85295james

:)

13. 85295james

hope that helps

14. monroe17

how'd you go from (x-2)^2+3(x-2)+4 to this.. x^2-4x+4+3(x-2)+4 ?

15. jim_thompson5910

(x-2)^2 (x-2)(x-2) x(x-2) - 2(x-2) x^2 - 2x - 2x + 4 x^2 - 4x + 4

16. jim_thompson5910

So (x-2)^2 is the same as x^2 - 4x + 4

17. jim_thompson5910

which makes (x-2)^2+3(x-2)+4 the same as x^2-4x+4+3(x-2)+4

18. monroe17

yea, yeah, I wasn't thinking.

19. jim_thompson5910

that's ok

20. monroe17

lol :)

21. monroe17

I'm gonna try to do this one by myself.. and i'll report back what answer I get lol.. (x^2+3x+4)^2+3(x^2+3x+4)+4...

22. jim_thompson5910

alright

23. monroe17

is it simplified to.. x^4+3x^2+8+3x^2+9x+12+4? I know that not all the way simplified.. I just wanna know if im sorta on the right track?

24. monroe17

x^4+3x^2+16+3x^2+9x+12+4..

25. jim_thompson5910

no, you're supposed to have cubed terms in there

26. jim_thompson5910

keep in mind that (x^2+3x+4)^2 (x^2+3x+4)(x^2+3x+4) x^2(x^2+3x+4)+3x(x^2+3x+4)+4(x^2+3x+4)

27. monroe17

(x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?

28. jim_thompson5910

that is correct, so (x^2+3x+4)^2+3(x^2+3x+4)+4 becomes x^4+6x^3+17x^2+24x+16+3(x^2+3x+4)+4

29. monroe17

x^4+6x^3+20x^2+33x+24?

30. monroe17

+32*

31. jim_thompson5910

good catch

32. jim_thompson5910

x^4+6x^3+20x^2+33x+32 is the correct final answer

33. monroe17

my goodness! finally lol! Now, one final question... How do I find the domain of those... :/

34. jim_thompson5910

The domain of any polynomial is the set of all real numbers.

35. jim_thompson5910

since you can plug in any real number into a polynomial function and get a real number out.

36. jim_thompson5910

If f(x) and g(x) are both polynomial functions, then the following are also polynomial functions f(g(x)), g(f(x)), f(f(x)), and g(g(x)) or any combination of composition you can think of are all polynomial functions

37. jim_thompson5910

So the domain of each composite function is the set of all real numbers.

38. jim_thompson5910

Notice nowhere do we have to worry about a) dividing by zero or b) taking the square root of negative numbers. So there are no restrictions on x.

39. monroe17

so the domain for all four polynomials is (-inf,inf)?

40. jim_thompson5910

exactly

41. monroe17

phew.. I thought it was gonna b longer then that. lol! Thank you so much. so for any polynomial.. the domain will be (-inf,inf) because there are no restrictions?

42. jim_thompson5910

Yes, any number will do for the input of a polynomial. So the domain in interval notation for any polynomial is (-inf,inf)

43. jim_thompson5910

You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)

44. monroe17

okay, I gotta remember that :) thank you!

45. jim_thompson5910

you're welcome