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angelwings996

  • 3 years ago

Find all the zeros of the equation. x^4 - 6x^2 - 7x - 6 = 0

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  1. mathmate
    • 3 years ago
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    Using Descartes rule of signs, there is one positive root, one or three negative roots.

  2. angelwings996
    • 3 years ago
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    I'm sorry, I'm so confused. I don't really know what I am suppose to do or use to figure this problem out

  3. mathmate
    • 3 years ago
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    The factorization theorem says if there are rational roots, they would be the factors of 6, or you can try factors like (x+1), (x-1), (x-2), (x+2), (x-3),(x+3),(x-6),(x+6). Since it can be shown that x+1 and x-1 are not factors, there are two out of the remaining six are zeroes. Let f(x)=x^4 - 6x^2 - 7x - 6 , you can try f(2)=2^4-6*2^2-7*2-6=16-24-14-6=-28 But since f(-2)=16-24+14-6=0, we conclude that x=-2 is a zero. Continue trying f(3) [because -2*3=-6, the constant term to see if x=3 is a zero.

  4. angelwings996
    • 3 years ago
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    Oh I get it. So I am just plugging in numbers 1 - 6 both positive and negative to see which one(s) will equal zero, correct ?

  5. mathmate
    • 3 years ago
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    Yes, except that you only plug in factors of 6. So you can skip 4 and 5, because 4 and 5 are not factors of 6. This is true when the leading coefficient is 1 (i.e. 1x^4). If the leading coefficient is not 1, then we have to check all combinations of (ax\(\pm b)\), where a is one of the factors of the leading coefficient, and b is one of the factors of the constant term. Thank you for the medal.

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