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mathmateBest ResponseYou've already chosen the best response.1
Using Descartes rule of signs, there is one positive root, one or three negative roots.
 one year ago

angelwings996Best ResponseYou've already chosen the best response.0
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
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
The factorization theorem says if there are rational roots, they would be the factors of 6, or you can try factors like (x+1), (x1), (x2), (x+2), (x3),(x+3),(x6),(x+6). Since it can be shown that x+1 and x1 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)=2^46*2^27*26=1624146=28 But since f(2)=1624+146=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.
 one year ago

angelwings996Best ResponseYou've already chosen the best response.0
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 ?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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.
 one year ago
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