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Using Descartes rule of signs, there is one positive root, one or three negative roots.
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
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
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.
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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 ?
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.