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anonymous

  • 5 years ago

what are the roots of i) f(x)=X^4-6x^3+10x^2+2x-15 ii) f(x)=x^3+11x^2+ 31x+21

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  1. anonymous
    • 5 years ago
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    try synthetic division

  2. anonymous
    • 5 years ago
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    You should try to find roots according to the rational toot theorem. If you can find rational roots, you can then divide these factors out to simplify the polynomial. The aim is to find a couple of factors in the first one you can use to diminish the degree of the quartic to 2. The rational root theorem gives -1 and 3 as possibilities here for the first polynomial (technically, +/1{1,3,5,15}, but -1 and 3 will work).

  3. anonymous
    • 5 years ago
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    *+/-*

  4. anonymous
    • 5 years ago
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    You can use the theorem again for the second. The theorem says, if you have a polynomial \[P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\]then if P has rational roots, they will be of the form,\[x=\frac{p}{q}\]where p and q are coprime in Z and where the numerator is an integer factor of the constant term \[a_0\]and the denominator is an integer factor of \[a_n\].

  5. anonymous
    • 5 years ago
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    If you exhaust all combinations of this form, your polynomial has no rational roots.

  6. anonymous
    • 5 years ago
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    The possible factors to consider in the numerator for the second equation are the factors of 21 (last term) +/-{1,3,7,21} The possible factors to consider for your denominator are +/-{1} (the coefficient of your highest term is 1) So you consider all possible combinations of \[\frac{p}{q}\]where p is from the first list and q is from the second.

  7. anonymous
    • 5 years ago
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    Here, for the second, if you try -1, -3 and -7, you'll find each is a root.

  8. anonymous
    • 5 years ago
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    ^ second equation, I mean.

  9. anonymous
    • 5 years ago
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    hi

  10. anonymous
    • 5 years ago
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    hi

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