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yociyoci

  • 2 years ago

How to factor: x^5-6x^4+40x^2-13x-70 Please show me the steps!

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  1. ArkGoLucky
    • 2 years ago
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    There is a theorem thing that states that plus or minus the factors of the constant(70 in this case) divided by plus or minus the constant in front of the term with the largest power are all the possibilities of the factors of the polynomial. So just start crunching the numbers.

  2. ArkGoLucky
    • 2 years ago
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    try \[\pm1,2,5,7,10,35,70\]then use synthetic division or polynomial long division to divide out the factors

  3. yociyoci
    • 2 years ago
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    can you explain synthetic divsion please?

  4. ArkGoLucky
    • 2 years ago
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    let finish solving the problem and then I will explain

  5. yociyoci
    • 2 years ago
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    Then, can you please show the steps for doing the polynomial long division to divide out the factors

  6. ArkGoLucky
    • 2 years ago
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    there is only one integer factor: (x-2). |dw:1357004129003:dw|

  7. ArkGoLucky
    • 2 years ago
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    the other factor is (x-2)

  8. yociyoci
    • 2 years ago
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    What are the numbers of the thrid and forth row?

  9. yociyoci
    • 2 years ago
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    oh, I am beginning to see how synethetic divsion works

  10. yociyoci
    • 2 years ago
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    Please correct me if I am going to wrong way: x^5-6x^4+40x^2-13x-70 =(x-2)(x^4-4x^3-8x^2+24x+35) The next step would be to divide (x^4-4x^3-8x^2+24x+35) by 5 using synethetic divsion?

  11. LogicalApple
    • 2 years ago
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    The remaining factors are not rational, so the rational roots theorem would not assist us any further.

  12. yociyoci
    • 2 years ago
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    oh, no wonder it didn't work anymore

  13. ArkGoLucky
    • 2 years ago
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    so that's how you factor it

  14. yociyoci
    • 2 years ago
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    (x-2)(x^4-4x^3-8x^2+24x+35) This is the simplest form? Can't (x^4-4x^3-8x^2+24x+35) be factor too?

  15. LogicalApple
    • 2 years ago
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    Since the remaining roots are complex, you couldn't factor out a single term, you would have to somehow factor out a quadratic. I do not know any techniques that will do this.

  16. LogicalApple
    • 2 years ago
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    Not complex, irrational. But they exist as conjugates so you would have to factor out a quadratic instead of a single term.

  17. yociyoci
    • 2 years ago
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    ok, i will continue to try

  18. yociyoci
    • 2 years ago
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    I got it (x^4-4x^3-8x^2+24x+35)=(x^2-2x-7)(x^2-2x-5)

  19. yociyoci
    • 2 years ago
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    so when (x^5-6x^4+40x^2-13x-70) is factored, it will be (x-2)(x^2-2x-7)(x^2-2x-5) Is it correct?

  20. LogicalApple
    • 2 years ago
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    Yes. That's it !

  21. mathmate
    • 2 years ago
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    @yociyoci just curious, how did you factorize it?

  22. yociyoci
    • 2 years ago
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    Thank you so much everyone for helping me with factoring! I learnt so much from you guys!! Thanks!!

  23. yociyoci
    • 2 years ago
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    I plug in variables and did some algebra and trial and error

  24. LogicalApple
    • 2 years ago
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    o_o

  25. yociyoci
    • 2 years ago
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    ya, it was long...

  26. mathmate
    • 2 years ago
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    You can also use some facts as we usually do in quadratics. Assume (x^2+ax+b)(x^2+cx+d) then (a+c)=-4 ac+b+d=-8 bc+ad=24 bd=35 Hopefully this will help you eliminate cases quickly without having to multiply out the whole expression.

  27. yociyoci
    • 2 years ago
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    ya, I did something similar

  28. LogicalApple
    • 2 years ago
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    That's a really good technique.

  29. mathmate
    • 2 years ago
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    Great! Thank you for the enjoyable session!

  30. yociyoci
    • 2 years ago
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    Thank you very much, guys!! :)

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