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Callisto

  • 2 years ago

Polynomials #1 Question 41 It is given that f(3x) = 54x^3 -27x^2 +px +q. When f(x) is divided by (x-3), the remainder is 42. Find the remainder when f(x/3) is divided by x-9 *Note: I'm helping my sister but I'm in trouble too :|*

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  1. FoolForMath
    • 2 years ago
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    I thought someone was typing an answer.

  2. cwtan
    • 2 years ago
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    TT bluring....... btw i found -4.... I duno whether it is right or wrong...

  3. Callisto
    • 2 years ago
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    The answer is not -4 :|

  4. cwtan
    • 2 years ago
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    younger sister? it's complicated... LOL

  5. Callisto
    • 2 years ago
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    Younger sister...

  6. Ackhat
    • 2 years ago
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    p+q=6?

  7. Callisto
    • 2 years ago
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    Nope :| p+q=15

  8. cwtan
    • 2 years ago
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    Big trouble

  9. Ackhat
    • 2 years ago
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    oh yes 15

  10. Ackhat
    • 2 years ago
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    i know wait

  11. Ackhat
    • 2 years ago
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    :) why my answer is 42

  12. Callisto
    • 2 years ago
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    It is!!!!

  13. Ackhat
    • 2 years ago
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    ohhohohoh

  14. Ishaan94
    • 2 years ago
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    Lol this question is funny.

  15. Callisto
    • 2 years ago
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    How did you get that answer? @Ackhat

  16. ash2326
    • 2 years ago
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    We have \[f(3x)=54x^3-27x^2+px+q\] It can be written as \[f(3x)=2(3x)^3-3(3x)^2+\frac p 3 (3x) +q\] so \[f(x)=2x^3-3x^2+\frac p 3 x+q\] We are given that when f(x) is divided by (x-3) the remainder is 42 so Using remainder theorem \[f(3)=42\] Now you can find P+q from here, Next find \(f(\frac x 3)\) to find the remainder when f(x/3) is divided by (x-9) put x=9 in f(x/3)

  17. Ackhat
    • 2 years ago
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    f(x)=2x^3-3x^2+p/3x+q

  18. Callisto
    • 2 years ago
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    .... I've thought about it.....I swear......

  19. Ishaan94
    • 2 years ago
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    f(x)/(x-3)=k+42/(x-3) 3f(x/3)/(x-9)= k + 3*42/(x-9)

  20. Ishaan94
    • 2 years ago
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    :D

  21. Ackhat
    • 2 years ago
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    i simpy divide by x-3 and got p+q=15 and after by x-9 and got that the reminder is p+q+27

  22. Callisto
    • 2 years ago
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    Hold on... @Ishaan94 How does it work: 3f(x/3)/(x-9)= k + 3*42/(x-9)? @Ackhat Do you mean you divided the equation f(x) by (x-3) first, then divide f(x) by (x-9)? or ...?

  23. Ackhat
    • 2 years ago
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    f(x)/(x-3) f(x/3)/(x-9)

  24. Ishaan94
    • 2 years ago
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    \[\frac{f(x)}{x-3} = P(x) + \frac{42}{x-3}\]\[x = \frac{x}{3}\] \[\frac{3\cdot f\left(x/3\right)}{x-9} = P(x) + \frac{42\cdot 3}{x-9}\] I love my solution <3

  25. Callisto
    • 2 years ago
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    Okay, got it. My calculation mistake :| @Ackhat

  26. Callisto
    • 2 years ago
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    @Ishaan94 More explanation is appreciated :) (sorry... I'm stupid :| )

  27. FoolForMath
    • 2 years ago
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    Beautiful solution Ishaan.

  28. FoolForMath
    • 2 years ago
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    Calli: Division Algorithm

  29. Ishaan94
    • 2 years ago
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    I am not a good teacher, Sorry. What part you didn't understand callisto?

  30. Ishaan94
    • 2 years ago
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    Thank you very much foolformath

  31. Callisto
    • 2 years ago
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    \[\frac{3\cdot f\left(x/3\right)}{x-9} = P(x) + \frac{42\cdot 3}{x-9}\] ^ don't know where it comes..

  32. Ackhat
    • 2 years ago
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    Bezu

  33. Ackhat
    • 2 years ago
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    :)

  34. cwtan
    • 2 years ago
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    What a simple solution!!! Great job @Ishaan94 !!!!!

  35. FoolForMath
    • 2 years ago
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    Substitute \(x=\frac x3 \) in \[ \frac{f(x)}{x-3} = P(x) + \frac{42}{x-3} \]

  36. Callisto
    • 2 years ago
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    Oh... Got it!!!! Thanks!!! Lovely solution :)

  37. Ishaan94
    • 2 years ago
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    Okay. \[\large\frac{f\left(\frac{x}{3}\right)}{\frac{x}3-3} = P + \frac{42}{\frac x3-3}\]Where P is any quadratic polynomial. \[\large \implies \frac{f\left(\frac x3\right)}{\frac{x-9}3} = P + \frac{42}{\frac {x -9}3}\]

  38. FoolForMath
    • 2 years ago
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    No, it is beautiful :|

  39. Ishaan94
    • 2 years ago
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    What's 'Bezu'? @Akchat

  40. Ishaan94
    • 2 years ago
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    @Ackhat *

  41. Ackhat
    • 2 years ago
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    never mind

  42. Ackhat
    • 2 years ago
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    ok this is a theorem

  43. Callisto
    • 2 years ago
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    The most wonderful thing is that my sister understands it :) Once again, thank you everyone :)

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