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Sgstudent Group Title

Given that 5x^3 + 6x^2 -x + 2 = (x+1) (x-2) Q(x) + ax + b find the value of a and b. Question on Identities :)

  • 2 years ago
  • 2 years ago

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  1. Ishaan94 Group Title
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    Q(x) is an unknown polynomial?

    • 2 years ago
  2. Sgstudent Group Title
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    yea

    • 2 years ago
  3. nphuongsun93 Group Title
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    then... a = -1, b = 2?...dont think it should be this simple

    • 2 years ago
  4. Sgstudent Group Title
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    that isnt the answer lol

    • 2 years ago
  5. Sgstudent Group Title
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    answer is a = 20 b = 24

    • 2 years ago
  6. Sgstudent Group Title
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    but i cant get it :(

    • 2 years ago
  7. Ishaan94 Group Title
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    if Q(x) = 5x-9 ?

    • 2 years ago
  8. Sgstudent Group Title
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    o..o maybe you could work out the full solution and see if it matches my answer :P

    • 2 years ago
  9. Sgstudent Group Title
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    Thank you.

    • 2 years ago
  10. Ishaan94 Group Title
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    \[ (x+​1)*​(x+​2)*​(5*​x-​9) = 5x^3 + 6x^2 -17x - 18 + 16x +20=5x^3 + 6x^2 -x + 2\]

    • 2 years ago
  11. Ishaan94 Group Title
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    @Sgstudent

    • 2 years ago
  12. Sgstudent Group Title
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    oo then how should i carry on

    • 2 years ago
  13. Ishaan94 Group Title
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    hmm a is 16 and b is 20 for me

    • 2 years ago
  14. Sgstudent Group Title
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    check for any careless mistake? :O

    • 2 years ago
  15. Ishaan94 Group Title
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    no i don't think so

    • 2 years ago
  16. Sgstudent Group Title
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    can you upload the picture of your working?

    • 2 years ago
  17. Ishaan94 Group Title
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    i didn't do it on my notebook, in head.

    • 2 years ago
  18. Ishaan94 Group Title
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    \[(x+​1)(x+​2)(5x−​9) +ax + b = 5x^3+6x^2−17x−18+ ax+b\]

    • 2 years ago
  19. cwrw238 Group Title
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    i did it also and got same result as ishaan

    • 2 years ago
  20. Sgstudent Group Title
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    i need the working :(

    • 2 years ago
  21. Sgstudent Group Title
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    or rather simple explanation towards the steps

    • 2 years ago
  22. cwrw238 Group Title
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    equating coefficients of x gives -1 = 10 - 27 + a a = 27-11 = 16

    • 2 years ago
  23. Ishaan94 Group Title
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    use q(x) as px +q, suggested by you earlier if you really need to show working

    • 2 years ago
  24. Sgstudent Group Title
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    can i let Q(x) = px + q will i be able to get the answer?

    • 2 years ago
  25. Sgstudent Group Title
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    oh but i cant seem to get the answer somehow :(

    • 2 years ago
  26. Sgstudent Group Title
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    i get decimal numbers :(

    • 2 years ago
  27. Ishaan94 Group Title
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    show me your working i will try to guide you

    • 2 years ago
  28. Sgstudent Group Title
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    kk i try again

    • 2 years ago
  29. Ishaan94 Group Title
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    well you can get multiple answers, maybe.

    • 2 years ago
  30. Sgstudent Group Title
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    only a and b

    • 2 years ago
  31. Sgstudent Group Title
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    (x+1) (x-2) *********

    • 2 years ago
  32. Sgstudent Group Title
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    Ishaan i edited the question

    • 2 years ago
  33. Sgstudent Group Title
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    sry about it

    • 2 years ago
  34. cwrw238 Group Title
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    i can see how ishaan got p = 5 because of the 5x^3 but the q = -9 is not obvious = 5x^3 + 15x^2 + 10x + qx^2 + 3qx + 2q + ax + b = 5x^3 + (15 + q)x^2 + (10 + 3q + a)x + 2q + b compare coefficients with 5x^3 + 6x^2 - x + 2 of x^2: q + 15 = 6 so q = -9 of x: 10 + a + 3q = 10 + a - 27 = a - 17 and a -17 = -1 so a = 16 of x^0 : 2q + b = 2 so -18 + b = 2 so b = 20 a=16, b = 20

    • 2 years ago
  35. Sgstudent Group Title
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    Thanks :)

    • 2 years ago
  36. cwrw238 Group Title
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    yw

    • 2 years ago
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