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monroe17 Group Title

((x^2+3x+4)-2) simplifies to -2x^2-6x-8

  • 2 years ago
  • 2 years ago

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  1. monroe17 Group Title
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    fog.. when f(x)=x-2 g(x)=x^2+3x+4 =((x^2+3x+4)-2) = -2x^2-6x-8

    • 2 years ago
  2. jim_thompson5910 Group Title
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    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

    • 2 years ago
  3. jim_thompson5910 Group Title
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    You're not multiplying, you're subtracting in this case

    • 2 years ago
  4. monroe17 Group Title
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    oh.. so I don't distribute the -2?

    • 2 years ago
  5. monroe17 Group Title
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    so for fof.. would is be.. (x-2)-2=x-4?

    • 2 years ago
  6. jim_thompson5910 Group Title
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    no, you're not distributing since it's originally x-2 and not -2x

    • 2 years ago
  7. jim_thompson5910 Group Title
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    yes, fof is x-4

    • 2 years ago
  8. 85295james Group Title
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    no the roots is X=-1 or -2

    • 2 years ago
  9. monroe17 Group Title
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    what about ... (x-2)^2+3(x-2)+4?

    • 2 years ago
  10. jim_thompson5910 Group Title
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    (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

    • 2 years ago
  11. 85295james Group Title
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    the polynomial discriminant: =1

    • 2 years ago
  12. 85295james Group Title
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    :)

    • 2 years ago
  13. 85295james Group Title
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    hope that helps

    • 2 years ago
  14. monroe17 Group Title
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    how'd you go from (x-2)^2+3(x-2)+4 to this.. x^2-4x+4+3(x-2)+4 ?

    • 2 years ago
  15. jim_thompson5910 Group Title
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    (x-2)^2 (x-2)(x-2) x(x-2) - 2(x-2) x^2 - 2x - 2x + 4 x^2 - 4x + 4

    • 2 years ago
  16. jim_thompson5910 Group Title
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    So (x-2)^2 is the same as x^2 - 4x + 4

    • 2 years ago
  17. jim_thompson5910 Group Title
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    which makes (x-2)^2+3(x-2)+4 the same as x^2-4x+4+3(x-2)+4

    • 2 years ago
  18. monroe17 Group Title
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    yea, yeah, I wasn't thinking.

    • 2 years ago
  19. jim_thompson5910 Group Title
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    that's ok

    • 2 years ago
  20. monroe17 Group Title
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    lol :)

    • 2 years ago
  21. monroe17 Group Title
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    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...

    • 2 years ago
  22. jim_thompson5910 Group Title
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    alright

    • 2 years ago
  23. monroe17 Group Title
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    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?

    • 2 years ago
  24. monroe17 Group Title
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    x^4+3x^2+16+3x^2+9x+12+4..

    • 2 years ago
  25. jim_thompson5910 Group Title
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    no, you're supposed to have cubed terms in there

    • 2 years ago
  26. jim_thompson5910 Group Title
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    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)

    • 2 years ago
  27. monroe17 Group Title
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    (x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?

    • 2 years ago
  28. jim_thompson5910 Group Title
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    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

    • 2 years ago
  29. monroe17 Group Title
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    x^4+6x^3+20x^2+33x+24?

    • 2 years ago
  30. monroe17 Group Title
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    +32*

    • 2 years ago
  31. jim_thompson5910 Group Title
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    good catch

    • 2 years ago
  32. jim_thompson5910 Group Title
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    x^4+6x^3+20x^2+33x+32 is the correct final answer

    • 2 years ago
  33. monroe17 Group Title
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    my goodness! finally lol! Now, one final question... How do I find the domain of those... :/

    • 2 years ago
  34. jim_thompson5910 Group Title
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    The domain of any polynomial is the set of all real numbers.

    • 2 years ago
  35. jim_thompson5910 Group Title
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    since you can plug in any real number into a polynomial function and get a real number out.

    • 2 years ago
  36. jim_thompson5910 Group Title
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    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

    • 2 years ago
  37. jim_thompson5910 Group Title
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    So the domain of each composite function is the set of all real numbers.

    • 2 years ago
  38. jim_thompson5910 Group Title
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    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.

    • 2 years ago
  39. monroe17 Group Title
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    so the domain for all four polynomials is (-inf,inf)?

    • 2 years ago
  40. jim_thompson5910 Group Title
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    exactly

    • 2 years ago
  41. monroe17 Group Title
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    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?

    • 2 years ago
  42. jim_thompson5910 Group Title
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    Yes, any number will do for the input of a polynomial. So the domain in interval notation for any polynomial is (-inf,inf)

    • 2 years ago
  43. jim_thompson5910 Group Title
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    You'll only run into domain restrictions if you deal with division of variables or having variables in square roots (or logs)

    • 2 years ago
  44. monroe17 Group Title
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    okay, I gotta remember that :) thank you!

    • 2 years ago
  45. jim_thompson5910 Group Title
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    you're welcome

    • 2 years ago
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