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kirk.freedman

Find all the zeroes of each equation. x^3 – x^2 + 5x – 5 = 0

  • one year ago
  • one year ago

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  1. kirk.freedman
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    Someone please help, how do I do this?

    • one year ago
  2. RadEn
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    the sum of all coefficients be 0 (1-1+5-5=0) it means x=1 is the root or (x-1) is one of the factor that polynomial find the other factor by using synthetic division, then for zeroes take all factors be 0 and solve for x

    • one year ago
  3. kirk.freedman
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    So I got x^2+5 When I used synthetic division. What now?

    • one year ago
  4. sirm3d
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    equate it to zero and use the quadratic formula.

    • one year ago
  5. kirk.freedman
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    Oh wait so if I have x^2+5=0 then x=i sqrt 5? That's one of my answers?

    • one year ago
  6. kirk.freedman
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    These Are the possible answers btw |dw:1354664328662:dw|

    • one year ago
  7. RadEn
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    just to check the value of D(discrimininant) with D=b^2-4ac D=0^2-4*1*5=-20 it means no solution for x real, but if u want representing x to in complex number use the quadratic formula : x=[-b +- sqrt(b^2-4ac)]/(2a)

    • one year ago
  8. kirk.freedman
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    What is a,b, and c for the quadratic formula?

    • one year ago
  9. kirk.freedman
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    is it 1,1, and 5?

    • one year ago
  10. RadEn
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    actually the general form a quadratic equation is ax^2 + bx + c = 0 so given x^2 + 5 = 0 it means a=1, b=0, c=5

    • one year ago
  11. kirk.freedman
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    But the original formula is x^3 – x^2 + 5x – 5 = 0 wouldn't b be 5 since it's 5x?? I'm confused there.

    • one year ago
  12. kirk.freedman
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    Where do you get the x^2+5=0 formula?

    • one year ago
  13. kirk.freedman
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    Oh nvm, I got it. Thanks for the help!

    • one year ago
  14. sirm3d
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    you have all the answers. your answer should be A.

    • one year ago
  15. kirk.freedman
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    Yes, that's what I put and got it right:) Thanks

    • one year ago
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