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apple_pi

  • 3 years ago

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

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  1. ParthKohli
    • 3 years ago
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    Use the rational roots test?

  2. apple_pi
    • 3 years ago
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    try it

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

  4. ParthKohli
    • 3 years ago
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    Checking all of them. Gimme a sec.

  5. hartnn
    • 3 years ago
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    forget 1,-1

  6. ParthKohli
    • 3 years ago
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    Yup, not 1 and -1.

  7. ParthKohli
    • 3 years ago
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    not 5.

  8. hartnn
    • 3 years ago
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    no rational roots

  9. apple_pi
    • 3 years ago
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    yeah, but it can still be factorised

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

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

  12. ParthKohli
    • 3 years ago
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    Yeah, we can.

  13. hartnn
    • 3 years ago
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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

  14. hartnn
    • 3 years ago
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    like bd=-5

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

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

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