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Math Help - Rational Root Theorem....

  1. #1
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    Smile Rational Root Theorem....

    I need help in figuring out the roots of this weird looking polynomial...

    Q1) Considering the polynomial...P(x) = x^4 - 4x^3 + 2x^2 + 4x + 4
    Find all the Rational Roots of this polynomial...?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by vikramtiwari View Post
    I need help in figuring out the roots of this weird looking polynomial...

    Q1) Considering the polynomial...P(x) = x^4 - 4x^3 + 2x^2 + 4x + 4
    Find all the Rational Roots of this polynomial...?
    Do you kow the rules...its \frac{ps}{qs}

    where p's are factors of the constant and q's are the factors of the coefficient of the highest order term?

    Do you need more help so in this case q=1 and p=4

    q factors{1} p factors{1,2,4}
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  3. #3
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    Thanks you Sir!

    But they are asking for the rational numbers, I guess...so that will 1, 1/2 and 1/4...so like do we just have to consider '1' as the rational number for this polynomials...??
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by vikramtiwari View Post
    Thanks you Sir!

    But they are asking for the rational numbers, I guess...so that will 1, 1/2 and 1/4...so like do we just have to consider '1' as the rational number for this polynomials...??
    You put the wrong numbers over the wrong numbers but good job...you almost got it!
    Since it is constant over coefficient the possibilites are just the p factors over the q factors or in other words {\frac{1}{1},\frac{2}{1},\frac{4}{1}} which is the same as {1,4,2}

    NOw to find which are roots put them in the original function!
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  5. #5
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    Hello,

    I don't see that 1 is a solution...

    And yes, integers are considered as rationals
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  6. #6
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    I am sorry for mistyping it...but yeah none of them satisfy the conditions...so the conclusion in that there is no rational number which would make this polynomial equal to zero..
    So does that mean, 1,2,4 are the possible rational roots but its possible that none of them would satisfy the condition?

    Thanks once again!
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  7. #7
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    Quote Originally Posted by vikramtiwari View Post
    So does that mean, 1,2,4 are the possible rational roots but its possible that none of them would satisfy the condition?
    "possible rational roots" should be a zero of the polynomial. Which condition are you talking about?

    @Mathstud28: What is the method you are talking about?
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  8. #8
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    I am sorry for misunderstanding again! For condition, I meant for the polynomial...when we plug all those possible rational numbers(1,4,2) we do not get zero in the polynomial equation...so does that mean there are no rational numbers which will satisfy this polynomial, like someone said at the first response above??

    Thank you!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    "possible rational roots" should be a zero of the polynomial. Which condition are you talking about?

    @Mathstud28: What is the method you are talking about?
    What do you mean what do I mean...it is the rational root theorem...or as it has been affectionately dubbed the "ps and qs"?
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  10. #10
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    yeah in fact that function has no real roots at all
    i checked it out on a graping calculator
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