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Math Help - find the zeros

  1. #1
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    find the zeros

    Suppose you know that 1/3 and -2 are zeros of 3x^4-x^3-21x^2-11x+6
    Find all the zeros!
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  2. #2
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    Hint: 1/3 and -2 are zero of (3x-1)(x+2).

    You need the Remainder Theorem.

    How's your Synthetic Division? Long division?
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  3. #3
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    not really good
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  4. #4
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    It's time to practice.

    Every time you find a zero, use it to deflate the polynomial to one degree lower. Eventually, you'll get it down to quadratic and the last two are easy.
    Last edited by TKHunny; May 13th 2008 at 05:44 AM.
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  5. #5
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    You could also use Vieta's Formulas and get the system
    \frac13-2+r_3+r_4=\frac13
    -\frac23r_3r_4=2
    and solve it simultaneously for the other roots. You should get the same quadratic (I'm bad at synthetic division too).
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  6. #6
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    Hello,

    Quote Originally Posted by gumi View Post
    Suppose you know that 1/3 and -2 are zeros of 3x^4-x^3-21x^2-11x+6
    Find all the zeros!
    I'd do it this way :

    \frac 13 and -2 are zeros of 3x^4-x^3-21x^2-11x+6

    Hence 3x^4-x^3-21x^2-11x+6=(x-\frac 13)(x+2)P(x)

    where P is a polynomial of degree 2.

    We can rewrite this way :

    \begin{aligned} 3x^4-x^3-21x^2-11x+6 &=3(x-\frac 13)(x+2)Q(x) \\<br />
&=(3x-1)(x+2)Q(x) \end{aligned}

    Q(x)=\frac{P(x)}{3}, but we don't care.

    Q(x)=ax^2+bx+c

    Since the first term is 3x^4, a=1 because we would have 3x*x*ax˛ as the dominating term.

    Hence Q(x)=x^2+bx+c

    Let's study the constant term :
    -1*2*c=6

    -> c=-3


    Thus Q(x)=x^2+bx-3

    Now, you can develop...
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