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Polynomials - Polynomial Equations

  1. Aug 8, 2007 #1
    Trying to ask my method of doing the question is correct or not? Try check is there any mistake please? I am beginner.
    The Question
    Find the real solutions to each of the following questions.
    (a) X³+X²-10X+8=0


    My attempt at a solution

    X³+X²-10X+8=0
    f(1)=1³+1²-10(1)+8
    f(1)=0
    X=1
    X-1=0 (one of the factor)

    X³+X²-10X+8=(X-1)(X²+kX-8)
    Compare the coefficient X²
    X²=kX²-X²
    1=k-1
    2=k

    X²+X-8=(X-2)(X+4)
    The factors are (X-1)(X-2)(X+4)
    **(From here start my working wonder correct or wrong)***
    1st Method of answering
    (X-1)(X-2)(X+4)=0
    X-2=0, X-1=0, X+4=0
    X=2, X=1, X=-4 #
    2nd Method of answering
    f(X)=(X-1)(X-2)(X+4)
    f(X)=0
    (X-1)(X-2)(X+4)=0
    X-2=0, X-1=0, X+4=0
    X=2, X=1, X=-4 #

    The 1st Method of answering or 2nd is correct?
     
    Last edited: Aug 8, 2007
  2. jcsd
  3. Aug 8, 2007 #2

    malawi_glenn

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    X²+x-8=(x-2)(x+4) ?

    You should try to learn to perform division of polynomials.
     
    Last edited: Aug 8, 2007
  4. Aug 8, 2007 #3

    symbolipoint

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    TheDanny, exactly what course are you studying?
     
  5. Aug 8, 2007 #4
    Both methods are correct, it's just implied the f(x) = 0 = (x-1)(x-2)(x+4)
     
  6. Aug 8, 2007 #5

    HallsofIvy

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    Yes, it is true that f(1)= 0 and so x= 1 is a root and x-1 is a factor of x3+ x2- 10x+ 8

    ?? I was under the impression that -2x+ 4x= 2x, not X. This can't possibly be true. Did you consider the possibility that x2+x-8= 0 has NO rational roots? If that is true, then you might consider either completing the square or using the quadratic equation to solve x2+ x- 8= 0.

    [/quote]The factors are (X-1)(X-2)(X+4)[/quote]
    NO, they aren't. That's your error

     
    Last edited: Aug 9, 2007
  7. Aug 8, 2007 #6

    learningphysics

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    He accidentally wrote X²+x-8 instead of X²+2x-8. Other than that everything is correct.

    TheDanny, your solution is fine (except for leaving out the 2 in the polynomial). Both ways of answering seem fine to me.
     
  8. Aug 8, 2007 #7
    Yeah i found that i accidentally wrote X²+x-8 without the 2 inside.
    HallsofIvy, maybe you right over here that
    X²+2X-8=(X-2)(X+4), or i should write

    X²+2X-8=0
    (X-2)(X+4)=0?

    I study Mathematics T that is A-level or Pre-U i think. Higher School Certificate (HSC). The HSC was the precursor to the GCE A levels in the UK, and is still the name of the pre-university examination in some states in Australia.
    Future Maths is hard or? I wonder my teacher tell me that not all can learn only some who is talented only. I will need all your help next time, thanks.
     
  9. Aug 8, 2007 #8

    HallsofIvy

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    Don't you just hate when that happens?

    Here I was thinking "It's so hot, I'll have a nice gin and tonic before I turn in. Next thing I know, it's 8:00 AM. I'd probably better not have a gin and tonic now!

    Yes, if x2+ 2x- 8= 0 then (x-2)(x+4)= 0 and the solutions are x= 2 and x= -4. You can, op, of course, check that by just doing the arithmetic:
    22[/aup]+ 2(2)-8= 4+ 4- 8= 0 and (-4)2+ 2(-4)+ 8= 16- 8-8= 0.
     
  10. Aug 8, 2007 #9
    hmm

    HallsofIvy,
    this is mine error?
    [/quote]The factors are (X-1)(X-2)(X+4)[/quote]
    NO, they aren't. That's your error


    Should it be The factors are (X-1),(X-2),(X+4)? put commas between each factor?
     
  11. Aug 9, 2007 #10
    I think so -- If you were to list the factors of 12 you wouldn't say 1*2*3*4*6*12, you'd say 1,2,3,4,6,12.
     
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