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monroe17

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

  • one year ago
  • one year ago

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  1. monroe17
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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

    • one year ago
  2. jim_thompson5910
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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

    • one year ago
  3. jim_thompson5910
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    You're not multiplying, you're subtracting in this case

    • one year ago
  4. monroe17
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    oh.. so I don't distribute the -2?

    • one year ago
  5. monroe17
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    so for fof.. would is be.. (x-2)-2=x-4?

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

    • one year ago
  7. jim_thompson5910
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    yes, fof is x-4

    • one year ago
  8. 85295james
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    no the roots is X=-1 or -2

    • one year ago
  9. monroe17
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    what about ... (x-2)^2+3(x-2)+4?

    • one year ago
  10. jim_thompson5910
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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

    • one year ago
  11. 85295james
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    the polynomial discriminant: =1

    • one year ago
  12. 85295james
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    :)

    • one year ago
  13. 85295james
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    hope that helps

    • one year ago
  14. monroe17
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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 ?

    • one year ago
  15. jim_thompson5910
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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

    • one year ago
  16. jim_thompson5910
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    So (x-2)^2 is the same as x^2 - 4x + 4

    • one year ago
  17. jim_thompson5910
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    which makes (x-2)^2+3(x-2)+4 the same as x^2-4x+4+3(x-2)+4

    • one year ago
  18. monroe17
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    yea, yeah, I wasn't thinking.

    • one year ago
  19. jim_thompson5910
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    that's ok

    • one year ago
  20. monroe17
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    lol :)

    • one year ago
  21. monroe17
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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...

    • one year ago
  22. jim_thompson5910
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    alright

    • one year ago
  23. monroe17
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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?

    • one year ago
  24. monroe17
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    x^4+3x^2+16+3x^2+9x+12+4..

    • one year ago
  25. jim_thompson5910
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    no, you're supposed to have cubed terms in there

    • one year ago
  26. jim_thompson5910
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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)

    • one year ago
  27. monroe17
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    (x^2+3x+4)^2=x^4+6x^3+17x^2+24x+16?

    • one year ago
  28. jim_thompson5910
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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

    • one year ago
  29. monroe17
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    x^4+6x^3+20x^2+33x+24?

    • one year ago
  30. monroe17
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    +32*

    • one year ago
  31. jim_thompson5910
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    good catch

    • one year ago
  32. jim_thompson5910
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    x^4+6x^3+20x^2+33x+32 is the correct final answer

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

    • one year ago
  34. jim_thompson5910
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    The domain of any polynomial is the set of all real numbers.

    • one year ago
  35. jim_thompson5910
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    since you can plug in any real number into a polynomial function and get a real number out.

    • one year ago
  36. jim_thompson5910
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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

    • one year ago
  37. jim_thompson5910
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    So the domain of each composite function is the set of all real numbers.

    • one year ago
  38. jim_thompson5910
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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.

    • one year ago
  39. monroe17
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    so the domain for all four polynomials is (-inf,inf)?

    • one year ago
  40. jim_thompson5910
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    exactly

    • one year ago
  41. monroe17
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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?

    • one year ago
  42. jim_thompson5910
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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)

    • one year ago
  43. jim_thompson5910
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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)

    • one year ago
  44. monroe17
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    okay, I gotta remember that :) thank you!

    • one year ago
  45. jim_thompson5910
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    you're welcome

    • one year ago
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