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apple_pi Group Title

Factor: x^4-2x^3+x^2+8x-5 Try it manually...how is it done?

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Use the rational roots test?

    • 2 years ago
  2. apple_pi Group Title
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    try it

    • 2 years ago
  3. ParthKohli Group Title
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    Okay. The possible rational roots are: 5 -5 1 -1

    • 2 years ago
  4. ParthKohli Group Title
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    Checking all of them. Gimme a sec.

    • 2 years ago
  5. hartnn Group Title
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    forget 1,-1

    • 2 years ago
  6. ParthKohli Group Title
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    Yup, not 1 and -1.

    • 2 years ago
  7. ParthKohli Group Title
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    not 5.

    • 2 years ago
  8. hartnn Group Title
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    no rational roots

    • 2 years ago
  9. apple_pi Group Title
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    yeah, but it can still be factorised

    • 2 years ago
  10. Algebraic! Group Title
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    into two quadratics... what's the point here?

    • 2 years ago
  11. apple_pi Group Title
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    Well...you're factoring it I guess you could get the roots But how do you get to the quadratics?

    • 2 years ago
  12. ParthKohli Group Title
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    Yeah, we can.

    • 2 years ago
  13. hartnn Group Title
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    i may suggest a way, (x^2+ax+b)(x^2+cx+d) = x^4-2x^3+x^2+8x-5 and compare the co-efficients, 4 unknowns

    • 2 years ago
  14. hartnn Group Title
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    like bd=-5

    • 2 years ago
  15. ParthKohli Group Title
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    Like for example, we have \(\sqrt2{}\) as a root, then a factor would be \(\rm x^2 - 2\).

    • 2 years ago
  16. AbhimanyuPudi Group Title
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    Well, @hartnn 's way is very good to get the quadratics

    • 2 years ago
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